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