# 7.4: Homework- u-substitution

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1. As a review of the chain rule from derivatives, find $$\frac{d}{dx} e^{-x^2 + 1}$$.
$$2x e^{x^2 + 1}$$
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2. Read through section 6A again and then read through section 6B.
3. Watch the Khan Academy video u-substitution.
4. Compute the following indefinite integral using the method of $$u$$-substitution.

$$\int (5x^4 + 2x) e^{x^5 + x^2}dx$$

$$e^{x^5 + x^2} + C$$
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5. Watch another example of $$u$$-substitution: u-substitution 2.
6. Compute the following indefinite integral using the method of $$u$$-substitution.

$$\int \frac{2x}{x^2 - 5}dx$$

$$\ln(x^2 - 5) + C$$
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7. 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.
1. $$\int e^{-3x}dx$$.
$$\frac{1}{-3} e^{-3x} + C$$
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2. $$\int \left( \frac{1}{2}x - 1 \right)^4dx$$
$$.4 \left( \frac{1}{2} x - 1 \right)^5 + C$$
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.

3. $$\int_0^2 (5x - 3)^3dx$$.
$$116$$
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4. $$\int \frac{1}{7x - 2}dx$$
$$\frac{1}{7} \ln(7x - 2) + C$$
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5. $$\int \frac{1}{\sqrt{0.5x - 1}}dx$$
$$4 \sqrt{0.5x - 1} + C$$
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8. Try some more $$u$$-substitution integrals.
1. $$\int (8x^3) (x^4 + 1)^2dx$$
1. We see that $$u = x^4 + 1$$ in this case.
2. We see $$\frac{du}{dx} = 4x^3$$, so $$dx = \frac{1}{4x^3} du$$.
3. 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*}

4. Integrating we have

$$2 \int u^2du = 2 \left( \frac{1}{3} u^3 \right ) + C = \frac{2}{3} u^3 + C$$

5. Substitution of $$u = x^4 + 1$$ yields

$$\frac{2}{3} (x^4 + 1)^3 + C$$

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2. $$\int (x^4 + 1)^2 (8x^3)dx$$
This is exactly the same as the previous problem, just written a different way. No need to redo work.
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3. $$\int (3x^2 - 1) e^{x^3 - x}dx$$
1. We see that $$u = x^3 - x$$ in this case.
2. We see $$\frac{du}{dx} = 3x^2 - 1$$, so $$dx = \frac{1}{3x^2 - 1} du$$.
3. We have

\begin{align*} & = \int (3x^2 - 1) e^u \frac{1}{3x^2 - 1} du \\ & = \int e^udu \end{align*}

4. Integrating we have

$$\int e^udu = e^u + C$$

5. Substitution of $$u = x^3 - x$$ yields

$$e^{x^3 - x} + C$$

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4. $$\int (x^2 - \frac{1}{3}) e^{x^3 - x}dx$$
$$\frac{1}{3} e^{x^3 - x} + C$$
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5. $$\int (e^x+ x)^5 (e^x+1)dx$$
$$\frac{1}{6} (e^x + x)^6 + C$$
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6. $$\int \frac{2x}{x^2 - 1}dx$$
$$\ln(x^2 -1) + C$$
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7. $$\int e^x \sqrt{e^x + 1}dx$$
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8. $$\int \frac{\sin(x)}{\cos(x)}dx$$
$$-\ln(\cos(x)) + C$$
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9. $$\int \frac{\sqrt{ \ln(x) }}{x}dx$$
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10. $$\int \frac{x}{(x^2 - 5)^3}dx$$
1. We see that $$u = x^2 - 5$$ in this case.
2. We see $$\frac{du}{dx} = 2x$$, so $$dx = \frac{1}{2x} du$$.
3. 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*}

4. Integrating we have

$$\int \frac{1}{2}u^{-3}du = - \frac{1}{4} u^{-2} + C$$

5. Substitution of $$u = x^2 - 5$$ yields

$$- \frac{1}{4} (x^2 - 5)^{-2} + C$$

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11. $$\int \frac{\cos(x)}{\sqrt{\sin(x) + 1}}dx$$
$$2\sqrt{\sin(x) + 1} + C$$
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12. $$\int \frac{1}{x (2 \ln(x) + 1)^4}dx$$.
$$-\frac{1}{6(2 \ln(x) + 1)^3} + C$$
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13. $$\int (e^{5x} + 1)^9 (e^{5x} )dx$$.
$$\frac{1}{2} (e^{5x} + 1)^{10} + C$$
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