7.4: Homework- u-substitution
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As a review of the chain rule from derivatives, find \(\frac{d}{dx} e^{-x^2 + 1}\).
\(2x e^{x^2 + 1}\)ans
- Read through section 6A again and then read through section 6B.
- Watch the Khan Academy video u-substitution .
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Compute the following indefinite integral using the method of \(u\)-substitution.
\(\int (5x^4 + 2x) e^{x^5 + x^2}dx\)
\(e^{x^5 + x^2} + C\)ans - Watch another example of \(u\)-substitution: u-substitution 2.
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Compute the following indefinite integral using the method of \(u\)-substitution.
\(\int \frac{2x}{x^2 - 5}dx\)
\(\ln(x^2 - 5) + C\)ans -
Reread the part about the chain rule shortcut for \(u\)-substitution in chapter 6 of the online notes, and reread Example 6B.2. Then try the following problems.
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\(\int e^{-3x}dx\).
\(\frac{1}{-3} e^{-3x} + C\)ans
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\(\int \left( \frac{1}{2}x - 1 \right)^4dx\)
\(.4 \left( \frac{1}{2} x - 1 \right)^5 + C\)ans
.
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\(\int_0^2 (5x - 3)^3dx\).
\(116\)ans
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\(\int \frac{1}{7x - 2}dx\)
\(\frac{1}{7} \ln(7x - 2) + C\)ans
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\(\int \frac{1}{\sqrt{0.5x - 1}}dx\)
\(4 \sqrt{0.5x - 1} + C\)ans
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\(\int e^{-3x}dx\).
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Try some more \(u\)-substitution integrals.
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\(\int (8x^3) (x^4 + 1)^2dx\)
- We see that \(u = x^4 + 1\) in this case.
- We see \(\frac{du}{dx} = 4x^3\), so \(dx = \frac{1}{4x^3} du\).
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We have
\[\begin{align*} & = \int (8x^3) u^2 \frac{1}{4x^3} du \\ & = \int 8 u^2 \frac{1}{4}du \\ & = 2 \int u^2du \end{align*}\]
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Integrating we have
\(2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C\)
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Substitution of \(u = x^4 + 1\) yields
\(\frac{2}{3} (x^4 + 1)^3 + C\)
ans -
\(\int (x^4 + 1)^2 (8x^3)dx\)
This is exactly the same as the previous problem, just written a different way. No need to redo work.ans
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\(\int (3x^2 - 1) e^{x^3 - x}dx\)
- We see that \(u = x^3 - x\) in this case.
- We see \(\frac{du}{dx} = 3x^2 - 1\), so \(dx = \frac{1}{3x^2 - 1} du\).
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We have
\[\begin{align*} & = \int (3x^2 - 1) e^u \frac{1}{3x^2 - 1} du \\ & = \int e^udu \end{align*}\]
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Integrating we have
\(\int e^udu = e^u + C\)
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Substitution of \(u = x^3 - x\) yields
\(e^{x^3 - x} + C\)
ans -
\(\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx\)
\(\frac{1}{3} e^{x^3 - x} + C\)ans
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\(\int (e^x+ x)^5 (e^x+1)dx\)
\(\frac{1}{6} (e^x + x)^6 + C\)ans
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\(\int \frac{2x}{x^2 - 1}dx\)
\(\ln(x^2 -1) + C\)ans
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\(\int e^x \sqrt{e^x + 1}dx\)
ans
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\(\int \frac{\sin(x)}{\cos(x)}dx\)
\(-\ln(\cos(x)) + C\)ans
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\(\int \frac{\sqrt{ \ln(x) }}{x}dx\)
ans
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\(\int \frac{x}{(x^2 - 5)^3}dx\)
- We see that \(u = x^2 - 5\) in this case.
- We see \(\frac{du}{dx} = 2x\), so \(dx = \frac{1}{2x} du\).
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We have
\[\begin{align*} & = \int \frac{x}{u^3}\frac{1}{2x}du \\ & = \int \frac{1}{2u^3}du \\ & = \int \frac{1}{2}u^{-3}du \end{align*}\]
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Integrating we have
\(\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C\)
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Substitution of \(u = x^2 - 5\) yields
\(- \frac{1}{4} (x^2 - 5)^{-2} + C\)
ans -
\(\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx\)
\(2\sqrt{\sin(x) + 1} + C\)ans
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\(\int \frac{1}{x (2 \ln(x) + 1)^4}dx\).
\(-\frac{1}{6(2 \ln(x) + 1)^3} + C\)ans
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\(\int (e^{5x} + 1)^9 (e^{5x} )dx\).
\(\frac{1}{2} (e^{5x} + 1)^{10} + C\)ans
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\(\int (8x^3) (x^4 + 1)^2dx\)