Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

2.4E: Exercises

  • Page ID
    18548
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    Partial Fractions

    Use partial fraction decomposition (or a simpler technique) to express the rational function as a sum or difference of two or more simpler rational expressions.

    Exercise \(\PageIndex{1}\)

    \(\dfrac{1}{(x−3)(x−2)}\)

    Answer

    \( \frac{1}{\left(x - 3\right) \left(x - 2\right)} = - \frac{1}{x - 2} + \frac{1}{x - 3} \)

    Exercise \(\PageIndex{2}\)

    \(\dfrac{x^2+1}{x(x+1)(x+2)}\)

    Answer

    \(\dfrac{x^2+1}{x(x+1)(x+2)} \quad = \quad −\dfrac{2}{x+1}+\dfrac{5}{2(x+2)}+\dfrac{1}{2x}\)

    Exercise \(\PageIndex{3}\)

    \(\dfrac{1}{x^3−x}\)

    Hint

    \( x^{3} - x = x \left(x - 1\right) \left(x + 1\right) \)

    Answer

    \( \frac{1}{x^{3} - x} = \frac{1}{2 \left(x + 1\right)} + \frac{1}{2 \left(x - 1\right)} - \frac{1}{x} \)

    Exercise \(\PageIndex{4}\)

    \(\dfrac{3x+1}{x^2}\)

    Answer

    \(\dfrac{3x+1}{x^2} \quad = \quad \dfrac{1}{x^2}+\dfrac{3}{x}\)

    Exercise \(\PageIndex{5}\)

    \(\dfrac{3x^2}{x^2+1}\)

    Hint

    Use long division first.

    Answer

    \( \frac{3 x^{2}}{x^{2} + 1} = 3 - \frac{3}{x^{2} + 1} \)

    Exercise \(\PageIndex{6}\)

    \(\dfrac{2x^4}{x^2−2x}\)

    Answer

    \(\dfrac{2x^4}{x^2−2x} \quad = \quad 2x^2+4x+8+\dfrac{16}{x−2}\)

    Exercise \(\PageIndex{7}\)

    \(\dfrac{1}{(x−1)(x^2+1)}\)

    Answer

    \( \frac{1}{\left(x - 1\right) \left(x^{2} + 1\right)} = - \frac{x + 1}{2 \left(x^{2} + 1\right)} + \frac{1}{2 \left(x - 1\right)} \)

    Exercise \(\PageIndex{8}\)

    \(\dfrac{1}{x^2(x−1)}\)

    Answer

    \(\dfrac{1}{x^2(x−1)} \quad = \quad −\dfrac{1}{x^2}−\dfrac{1}{x}+\dfrac{1}{x−1}\)

    Exercise \(\PageIndex{9}\)

    \(\dfrac{x}{x^2−4}\)

    Answer

    \( \frac{x}{x^{2} - 4} = \frac{1}{2 \left(x + 2\right)} + \frac{1}{2 \left(x - 2\right)} \)

    Exercise \(\PageIndex{10}\)

    \(\dfrac{1}{x(x−1)(x−2)(x−3)}\)

    Answer

    \(\dfrac{1}{x(x−1)(x−2)(x−3)} \quad = \quad −\dfrac{1}{2(x−2)}+\dfrac{1}{2(x−1)}−\dfrac{1}{6x}+\dfrac{1}{6(x−3)}\)

    Exercise \(\PageIndex{11}\)

    \(\dfrac{1}{x^4−1}=\dfrac{1}{(x+1)(x−1)(x^2+1)}\)

    Answer

    \( \frac{3 x^{2}}{x^{3} - 1} = \frac{2 x + 1}{x^{2} + x + 1} + \frac{1}{x - 1} \)

    Exercise \(\PageIndex{12}\)

    \(\dfrac{3x^2}{x^3−1}=\dfrac{3x^2}{(x−1)(x^2+x+1)}\)

    Answer

    \(\dfrac{3x^2}{x^3−1} \quad = \quad \dfrac{1}{x−1}+\dfrac{2x+1}{x^2+x+1}\)

    Exercise \(\PageIndex{13}\)

    \(\dfrac{2x}{(x+2)^2}\)

    Answer

    \( \frac{2 x}{\left(x + 2\right)^{2}} = \frac{2}{x + 2} - \frac{4}{\left(x + 2\right)^{2}} \)

    Exercise \(\PageIndex{14}\)

    \(\dfrac{3x^4+x^3+20x^2+3x+31}{(x+1)(x^2+4)^2}\)

    Answer
    \(\dfrac{3x^4+x^3+20x^2+3x+31}{(x+1)(x^2+4)^2} \quad = \quad \dfrac{2}{x+1}+\dfrac{x}{x^2+4}−\dfrac{1}{(x^2+4)^2}\)

    In exercises 15 - 25, use the method of partial fractions to evaluate each of the following integrals.

    Exercise \(\PageIndex{15}\)

    \(\displaystyle ∫\frac{dx}{(x−3)(x−2)}\)

    Hint

    \( \frac{1}{\left(x - 3\right) \left(x - 2\right)} = - \frac{1}{x - 2} + \frac{1}{x - 3} \)

    Answer

    \(\displaystyle ∫\frac{dx}{(x−3)(x−2)} = \ln|x-3| - \ln|x-2| + C\)

    Exercise \(\PageIndex{16}\)

    \(\displaystyle ∫\frac{3x}{x^2+2x−8}\,dx\)

    Hint

    \( \frac{3 x}{x^{2} + 2 x - 8} = \frac{2}{x + 4} + \frac{1}{x - 2} \)

    Answer

    \(\displaystyle ∫\frac{3x}{x^2+2x−8}\,dx \quad = \quad −\ln|2−x|+2\ln|4+x|+C = \ln\left| \frac{(4+x)^2}{2-x} \right| + C\)

    Exercise \(\PageIndex{17}\)

    \(\displaystyle ∫\frac{dx}{x^3−x}\)

    Hint

    \( x^{3} - x = x \left(x - 1\right) \left(x + 1\right) \)

    Answer

    \(\displaystyle ∫\frac{dx}{x^3−x} = \frac{1}{2} \left( \ln|x+1| +\ln|x-1| -2\ln|x| \right) + C \)

    Exercise \(\PageIndex{18}\)

    \(\displaystyle ∫\frac{x}{x^2−4}\,dx\)

     

    Answer

    \(\displaystyle ∫\frac{x}{x^2−4}\,dx \quad = \quad \tfrac{1}{2}\ln|4−x^2|+C\)

    Exercise \(\PageIndex{19}\)

    \(\displaystyle ∫\frac{dx}{x(x−1)(x−2)(x−3)}\)

    Hint

    \( \frac{1}{x \left(x - 3\right) \left(x - 2\right) \left(x - 1\right)} = \frac{1}{2 \left(x - 1\right)} - \frac{1}{2 \left(x - 2\right)} + \frac{1}{6 \left(x - 3\right)} - \frac{1}{6 x} \)

    Answer

    \(\displaystyle ∫\frac{dx}{x(x−1)(x−2)(x−3)} = \frac{1}{6} \left( 3\ln|x-1| + 3\ln|x-2| + \ln|x-3| - \ln|x| \right) + C \)

    Exercise \(\PageIndex{20}\)

    \(\displaystyle ∫\frac{2x^2+4x+22}{x^2+2x+10}\,dx\)

    Answer

    \(\displaystyle ∫\frac{2x^2+4x+22}{x^2+2x+10}\,dx \quad = \quad 2\left(x+\tfrac{1}{3}\arctan\left(\frac{1+x}{3}\right)\right)+C\)

    Exercise \(\PageIndex{21}\)

    \(\displaystyle ∫\frac{dx}{x^2−5x+6}\)

    Hint

    \( x^{2} - 5 x + 6 = \left(x - 3\right) \left(x - 2\right) \)

    Answer

    \(\displaystyle ∫\frac{dx}{x^2−5x+6} = \ln|x-3|-\ln|x-2| +C\)

    Exercise \(\PageIndex{22}\)

    \(\displaystyle ∫\frac{2−x}{x^2+x}\,dx\)

    Hint

    \(\displaystyle \frac{2−x}{x^2+x} = \frac{-3}{x+1}+\frac{2}{x} \)

    Answer

    \(\displaystyle ∫\frac{2−x}{x^2+x}\,dx \quad = \quad 2\ln|x|−3\ln|1+x|+C = \ln\left| \frac{x^2}{(1+x)^3} \right|+C\)

    Exercise \(\PageIndex{23}\)

    \(\displaystyle ∫\frac{2}{x^2−x−6}\,dx\)

    Answer

    \(\displaystyle \frac{2}{x^2−x−6} = \frac{2\ln|x-3|}{5}-\frac{2\ln|x+2|}{5} + C\)

    Exercise \(\PageIndex{24}\)

    \(\displaystyle ∫\frac{dx}{x^3−2x^2−4x+8}\)

    Hint

    \( x^{3} - 2 x^{2} - 4 x + 8 = \left(x - 2\right)^{2} \left(x + 2\right) \)

    Answer

    \(\displaystyle ∫\frac{dx}{x^3−2x^2−4x+8} \quad = \quad \tfrac{1}{16}\left(−\frac{4}{−2+x}−\ln|−2+x|+\ln|2+x|\right)+C = \tfrac{1}{16}\left(−\frac{4}{−2+x}+\ln\left| \frac{x+2}{x-2} \right|\right)+C\)

    Exercise \(\PageIndex{25}\)

    \(\displaystyle ∫\frac{dx}{x^4−10x^2+9}\)

    Hint

    \( x^{4} - 10 x^{2} + 9 = \left(x - 3\right) \left(x - 1\right) \left(x + 1\right) \left(x + 3\right) \)

    Answer

    \(\displaystyle ∫\frac{dx}{x^4−10x^2+9} = \frac{1}{48} \left( \ln|x-3| - \ln|x+3| -3\ln|x-1|+3\ln|x+1|\right) + C \)

    In exercises 26 - 29, evaluate the integrals with irreducible quadratic factors in the denominators.

    Exercise \(\PageIndex{26}\)

    \(\displaystyle ∫\frac{2}{(x−4)(x^2+2x+6)}\,dx\)

    Answer

    \(\displaystyle ∫\frac{2}{(x−4)(x^2+2x+6)}\,dx \quad = \quad \tfrac{1}{30}(−2\sqrt{5}\arctan\left[\frac{1+x}{\sqrt{5}}\right]+2\ln|−4+x|−\ln|6+2x+x^2|)+C\)

    Exercise \(\PageIndex{27}\)

    \(\displaystyle ∫\frac{x^2}{x^3−x^2+4x−4}\,dx\)

    Hint

    \( x^{3} - x^{2} + 4 x - 4 = \left(x - 1\right) \left(x^{2} + 4\right) \)

    Answer

    \(\displaystyle ∫\frac{x^2}{x^3−x^2+4x−4}\ = \frac{1}{5} \left( \ln|x-1| +\ln \left(x^2+4\right) + 2 \arctan \frac{x}{2} \right) + C\)

    Exercise \(\PageIndex{28}\)

    \(\displaystyle ∫\frac{x^3+6x^2+3x+6}{x^3+2x^2}\,dx\)

    Hint

     \( x^{3} + 2 x^{2} = x^{2} \left(x + 2\right) \)

    Answer

    \(\displaystyle ∫\frac{x^3+6x^2+3x+6}{x^3+2x^2}\,dx \quad = \quad −\frac{3}{x}+4\ln|x+2|+x+C\)

    Exercise \(\PageIndex{29}\)

    \(\displaystyle ∫\frac{x}{(x−1)(x^2+2x+2)^2}\,dx\)

    Answer

    \( \displaystyle ∫\frac{x}{(x−1)(x^2+2x+2)^2} =  \frac{x^5}{5}+\frac{x^4}{4}+\frac{13x^3}{3}+\frac{21x^2}{2}+25x+25\ln|x-1| + C \)

    In exercises 30 - 32, use the method of partial fractions to evaluate the integrals.

    Exercise \(\PageIndex{30}\)

    \(\displaystyle ∫\frac{3x+4}{(x^2+4)(3−x)}\,dx\)

    Hint

    \( \frac{3 x + 4}{\left(- x + 3\right) \left(x^{2} + 4\right)} = \frac{x}{x^{2} + 4} - \frac{1}{x - 3} \)

    Answer

    \(\displaystyle ∫\frac{3x+4}{(x^2+4)(3−x)}\,dx \quad = \quad −\ln|3−x|+\tfrac{1}{2}\ln|x^2+4|+C\)

    Exercise \(\PageIndex{31}\)

    \(\displaystyle ∫\frac{2}{(x+2)^2(2−x)}\,dx\)

    Hint

    \( \displaystyle \frac{2}{\left(- x + 2\right) \left(x + 2\right)^{2}} = \frac{1}{8 \left(x + 2\right)} + \frac{1}{2 \left(x + 2\right)^{2}} - \frac{1}{8 \left(x - 2\right)} \)

    Answer

    \(\displaystyle ∫\frac{2}{(x+2)^2(2−x)} = -2\ln|x+2| -\frac{8}{x+2} +C \)

    Exercise \(\PageIndex{32}\)

    \(\displaystyle ∫\frac{3x+4}{x^3−2x−4}\,dx\) (Hint: Use the rational root theorem.)

    Hint

    \( \displaystyle \frac{3 x + 4}{x^{3} - 2 x - 4} = - \frac{x + 1}{x^{2} + 2 x + 2} + \frac{1}{x - 2} \)

    Answer

    \(\displaystyle ∫\frac{3x+4}{x^3−2x−4}\,dx \quad = \quad \ln|x−2|−\tfrac{1}{2}\ln|x^2+2x+2|+C\)

     

    In exercises 33 - 46, use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals.

    Exercise \(\PageIndex{33}\)

    \(\displaystyle ∫^1_0\frac{e^x}{36−e^{2x}}\,dx\) (Give the exact answer and the decimal equivalent. Round to five decimal places.)

     

    Answer

    \(\displaystyle ∫^1_0\frac{e^x}{36−e^{2x}} = \frac{1}{12} \left( \ln(e+6)+\ln(5)-\ln(7)-\ln(6-e) \right) \cong 0.05338 \)

    Exercise \(\PageIndex{34}\)

    \(\displaystyle ∫\frac{e^x\,dx}{e^{2x}−e^x}\,dx\)

    Hint

    \( \frac{e^x}{e^{2x}-e^x}=\frac{1}{e^x-1} \)

    Answer

    \( \displaystyle ∫\frac{e^x\,dx}{e^{2x}−e^x}\,dx = \quad −x+\ln|e^x-1|+C \)

    Exercise \(\PageIndex{35}\)

    \(\displaystyle ∫\frac{\sin x\,dx}{1−\cos^2x}\)

    Hint

    \(\displaystyle 1−\cos^2x = sin^2 x \)

    Answer

    \(\displaystyle ∫\frac{\sin x\,dx}{1−\cos^2x} = \frac{1}{2}\left( \ln| \cos x -1| - \ln|\cos(x)+1| \right) + C\)

    Exercise \(\PageIndex{36}\)

    \(\displaystyle ∫\frac{\sin x}{\cos^2 x+\cos x−6}\,dx\)

    Hint

    \( \displaystyle \cos^2 x+\cos x−6 = \left( \cos x-2 \right) \left(\cos x+3 \right) \)

    Answer

    \( \displaystyle ∫\frac{\sin x}{\cos^2 x+\cos x−6}\,dx \quad = \quad \tfrac{1}{5}\ln\left|\frac{\cos x+3}{\cos x−2}\right| + C\)

    Exercise \(\PageIndex{37}\)

    \(\displaystyle ∫\frac{1−\sqrt{x}}{1+\sqrt{x}}\,dx\)

    Hint

    After rationalizing the denominator, let u= 1-x.

    Answer

    \( \displaystyle ∫\frac{1−\sqrt{x}}{1+\sqrt{x}}\,dx = 4x^{1/2} - x - 4 \ln|x^{1/2}+1| + C \)

    Exercise \(\PageIndex{38}\)

    \(\displaystyle ∫\frac{dt}{(e^t−e^{−t})^2}\)

    Hint

    Put over a common denominator then use substitution.

    Answer

    \(\displaystyle ∫\frac{dt}{(e^t−e^{−t})^2} \quad = \quad \frac{1}{2−2e^{2t}}+C\)

    Exercise \(\PageIndex{39}\)

    \(\displaystyle ∫\frac{1+e^x}{1−e^x}\,dx\)

    Answer

    \(\displaystyle ∫\frac{1+e^x}{1−e^x}\,dx = -1 -\frac{2}{e^x-1} + C\)

    Exercise \(\PageIndex{40}\)

    \(\displaystyle ∫\frac{dx}{1+\sqrt{x+1}}\)

    Answer

    \(\displaystyle ∫\frac{dx}{1+\sqrt{x+1}} \quad = \quad 2\sqrt{1+x}−2\ln|1+\sqrt{1+x}|+C\)

    Exercise \(\PageIndex{41}\)

    \(\displaystyle ∫\frac{dx}{\sqrt{x}+\sqrt[4]{x}}\)

    Hint

    let \( u=x^{1/4} \)

    Answer

    \(\displaystyle ∫\frac{dx}{\sqrt{x}+\sqrt[4]{x}} = 2 \sqrt{x} -4x^{1/4}+4\ln(x^{1/4}+1) + C\)

    Exercise \(\PageIndex{42}\)

    \(\displaystyle ∫\frac{\cos x}{\sin x(1−\sin x)}\,dx\)

    Hint

    Let \( u = 1-\sin(x) \)

    Answer

    \(\displaystyle ∫\frac{\cos x}{\sin x(1−\sin x)}\,dx \quad = \quad \ln\left|\frac{\sin x}{1−\sin x}\right|+C\)

    Exercise \(\PageIndex{43}\)

    \(\displaystyle ∫\frac{e^x}{(e^{2x}−4)^2}\,dx\)

    Hint

    Let \( u=e^x \), then \( \frac{1}{(u^2-4)^2} = \frac{1}{32 \left(u + 2\right)} + \frac{1}{16 \left(u + 2\right)^{2}} - \frac{1}{32 \left(u - 2\right)} + \frac{1}{16 \left(u - 2\right)^{2}}\)

    Answer

    \( \displaystyle ∫ \frac{e^x}{(e^{2x}−4)^2}\,dx = - \frac{\ln \left(e^{x} - 2 \right)}{32} + \frac{\ln{\left(e^{x} + 2 \right)}}{32} - \frac{e^{x}}{8 e^{2 x} - 32} + C\)

    Exercise \(\PageIndex{44}\)

    \(\displaystyle ∫_1^2\frac{1}{x^2\sqrt{4−x^2}}\,dx\)

    Hint

    Perform a trig substitution.  Let \( x=2\sin(u) \)

    Answer

    \( ∫_1^2\frac{1}{x^2\sqrt{4−x^2}}\,dx = \quad \frac{\sqrt{3}}{4}\)

    Exercise \(\PageIndex{45}\)

    \(\displaystyle ∫\frac{1}{2+e^{−x}}\,dx\)

    Hint

    Get common denominator then use partial fraction.

    Answer:

    \( \int \frac{1}{2+e^{-x}}\,dx = \frac{x}{2} + \frac{\ln{\left(2 + e^{- x} \right)}}{2} + C\)

    Exercise \(\PageIndex{46}\)

    \(\displaystyle ∫\frac{1}{1+e^x}\,dx\)

    Hint

    Let \( u=1+e^x \)

    Answer

    \(\displaystyle ∫\frac{1}{1+e^x}\,dx \quad = \quad x−\ln(1+e^x)+C\)

    In exercises 47 - 48, use the given substitution to convert the integral to an integral of a rational function, then evaluate.

    Exercise \(\PageIndex{47}\)

    \(\displaystyle ∫\frac{1}{t−\sqrt[3]{t}}\,dt; \)

    Hint

    \( \quad t=x^3\)

    Answer

    \( \int \frac{1}{t-\sqrt[3]{t}}\,dt = \frac{3 \ln|t^{2/3}-1|}{2} + C \)

    Exercise \(\PageIndex{48}\)

    \(\displaystyle ∫\frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx; \)

    Hint

    \( \quad x=u^6 \)

    Answer

    \(\displaystyle ∫\frac{1}{\sqrt{x}+\sqrt[3]{x}}\,dx \quad = \quad 6x^{1/6}−3x^{1/3}+2\sqrt{x}−6\ln(1+x^{1/6})+C\)

     

    Exercise \(\PageIndex{49}\)

    Graph the curve \(y=\dfrac{x}{1+x}\) over the interval \([0,5]\). Then, find the area of the region bounded by the curve, the \(x\)-axis, and the line \(x=4\).

     

    Answer

    \( \int_0^4 \frac{x}{1+x} \, dx = 4-\ln(5) \)

    Exercise \(\PageIndex{50}\)

    Find the volume of the solid generated when the region bounded by \(y=\dfrac{1}{\sqrt{x(3−x)}}, \,y=0, \,x=1,\) and \(x=2\) is revolved about the \(x\)-axis.

     

    Answer

    \(V = \frac{4}{3}π \arctan h\,\left[\frac{1}{3}\right]=\frac{1}{3}π\ln 4 \, \text{units}^3\)

    Exercise \(\PageIndex{51}\)

    The velocity of a particle moving along a line is a function of time given by \(v(t)=\dfrac{88t^2}{t^2+1}.\) Find the distance that the particle has traveled after \(t=5\) sec.

    Hint

    \( \frac{88 t^{2}}{t^{2} + 1} = 88 - \frac{88}{t^{2} + 1}= 88(1-\frac{1}{t^2+1}) \)

    Answer

    \( \int_0^5 \frac{88t^2}{t^2+1}\,dx; = 440 - \arctan(5) = 319 \; \text{units}. \)

    In exercises 52 - 54, solve the initial-value problem for \(x\) as a function of \(t\).

    Exercise \(\PageIndex{52}\)

    \((t^2−7t+12)\dfrac{dx}{dt}=1,\quad t>4,\, x(5)=0\)

    Hint

    \( \frac{1}{t^{2} - 7 t + 12} = - \frac{1}{t - 3} + \frac{1}{t - 4} \)

    Answer

    \( x=−\ln|t−3|+\ln|t−4|+\ln 2 = \ln\left| \frac{2(t-4)}{t-3}\right|\)

    Exercise \(\PageIndex{53}\)

    \((t+5)\dfrac{dx}{dt}=x^2+1, \quad t>−5,\,x(1)=\tan 1\)

    Hint

    \( \int \frac{dt}{t+5} = \int \frac{dx}{x^2+1} \)

    Answer

    \( x(0) = 1- \ln(6) \)

    Exercise \(\PageIndex{54}\)

    \((2t^3−2t^2+t−1)\dfrac{dx}{dt}=3,\quad x(2)=0\)

    Hint

    \( \frac{1}{2 t^{3} - 2 t^{2} + t - 1} = - \frac{2 \left(t + 1\right)}{3 \left(2 t^{2} + 1\right)} + \frac{1}{3 \left(t - 1\right)} \)

    Answer

    \(x=\ln|t−1|−\sqrt{2}\arctan(\sqrt{2}t)−\frac{1}{2}\ln(t^2+\frac{1}{2})+\sqrt{2}\arctan(2\sqrt{2})+\frac{1}{2}\ln 4.5\)

    Exercise \(\PageIndex{55}\)

    Find the \(x\)-coordinate of the centroid of the area bounded by \(y(x^2−9)=1, \, y=0, \,x=4,\) and \(x=5.\) (Round the answer to two decimal places.)

     

    Answer

    Add texts here. Do not delete this text first.

    Exercise \(\PageIndex{56}\)

    Find the volume generated by revolving the area bounded by \(y=\dfrac{1}{x^3+7x^2+6x},\, x=1,\, x=7\), and \(y=0\) about the \(y\)-axis.

    Hint

    \( \frac{1}{x^{3} + 7 x^{2} + 6 x} = \frac{1}{30 \left(x + 6\right)} - \frac{1}{5 \left(x + 1\right)} + \frac{1}{6 x} \)

    Answer

    \(V = \frac{2}{5}π\ln\frac{28}{13} \, \text{units}^3\)

    Exercise \(\PageIndex{57}\)

    Find the area bounded by \(y=\dfrac{x−12}{x^2−8x−20}, \,y=0, \,x=2,\) and \(x=4\). (Round the answer to the nearest hundredth.)

    Hint

    \( \frac{x - 12}{x^{2} - 8 x - 20} = \frac{7}{6 \left(x + 2\right)} - \frac{1}{6 \left(x - 10\right)} \)

    Answer

    \( \int_2^4 \frac{x-12}{x^2-8x-20}\, dx = 0.52 \text{units.} \)

    Exercise \(\PageIndex{58}\)

    Evaluate the integral \(\displaystyle ∫\frac{dx}{x^3+1}.\)

    Hints

    \( \frac{1}{x^{3} + 1} = - \frac{x - 2}{3 \left(x^{2} - x + 1\right)} + \frac{1}{3 \left(x + 1\right)} \)

    \( x-2=\frac{1}{2}(2x-1)-\frac{3}{2} \; \text{then partial fraction. } \)

    Answer

    \(\displaystyle ∫\frac{dx}{x^3+1} \quad = \quad \frac{\arctan[\frac{−1+2x}{\sqrt{3}}]}{\sqrt{3}}+\frac{1}{3}\ln|1+x|−\frac{1}{6}\ln∣1−x+x^2∣+C\)

    For problems 59 - 62, use the substitutions \(\tan(\frac{x}{2})=t, \,dx=\dfrac{2}{1+t^2}\,dt, \, \sin x=\dfrac{2t}{1+t^2},\) and \(\cos x=\dfrac{1−t^2}{1+t^2}.\)

    Exercise \(\PageIndex{59}\)

    \(\displaystyle ∫\frac{dx}{3−5\sin x}\)

    Hint

    \( \sin x= \frac{2t}{1+t^2}\text{;} dx=\frac{2dt}{t^2+1} \text{; } t= \tan(\frac{x}{2}) \)

    Answer

    \( \displaystyle ∫\frac{dx}{3-5\sin x} = \frac{\ln|\tan(\frac{x}{2}-3| - \ln|3\tan\frac{x}{2}-1|}{4}+C \)

    Exercise \(\PageIndex{60}\)

    Find the area under the curve \(y=\dfrac{1}{1+\sin x}\) between \(x=0\) and \(x=π.\) (Assume the dimensions are in inches.)

    Hint

    \( \tan\frac{x}{2}=\frac{\sin(x)}{1+\cos(x)} \)

    Answer

    2.0 in.2

    Exercise \(\PageIndex{61}\)

    Given \(\tan(\frac{x}{2})=t,\) derive the formulas \(dx=\dfrac{2}{1+t^2}dt, \,\sin x=\dfrac{2t}{1+t^2}\), and \(\cos x=\dfrac{1−t^2}{1+t^2}.\)

     

    Answer

    \( \frac{d \tan \frac{x}{2} }{dx} = \frac{dt}{dx} =\frac{\sec^2 (\frac{x}{2}) }{2} \; \text{therefore} \; dx=\frac{2dt}{1+t^2} \)

    \( \sin(x)= 2 \sin\frac{x}{2} \cos\frac{x}{2} = \frac{2 \sin\frac{x}{2} \cos\frac{x}{2}}{\cos^2{x}+sin^2{x}} = ... \)

    Exercise \(\PageIndex{62}\)

    Evaluate \(\displaystyle ∫\frac{\sqrt[3]{x−8}}{x}\,dx.\)

    Hint

    Let \( u = \sqrt[3]{x-8}, \text{then} \; x=u^3+8 \)

    Answer

    \(\displaystyle ∫\frac{\sqrt[3]{x−8}}{x}\,dx \quad = \quad 3(−8+x)^{1/3}−2\sqrt{3}\arctan\left[\frac{−1+(−8+x)^{1/3}}{\sqrt{3}}\right]−2\ln\left[2+(−8+x)^{1/3}\right]+\ln\left[4−2(−8+x)^{1/3}+(−8+x)^{2/3}\right]+C\)