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This exercise gives you new problems to practice—ones requiring integration by parts.
Your in-class quiz will cover all the problems from the first three integration quizzes, together with these new problems.

In this exercise, all functions are assumed to have the required properties for a particular situation.
For example, any function in a denominator is assumed to be nonzero, any function inside a logarithm is assumed to be positive, and so on.


Integration by parts is the integration counterpart to the product rule for differentiation.

Let $u$ and $v$ both be functions of $x$, so that: $$\frac{d}{dx} uv = u\frac{dv}{dx} + v\frac{du}{dx}$$ Integrating both sides with respect to $\,x\,$ gives: $$\int \frac{d}{dx} uv\ dx = \int u\frac{dv}{dx}\ dx + \int v\frac{du}{dx}\ dx$$ This simplifies to: $$uv = \int u\,dv + \int v\,du$$ A bit of rearrangement gives the INTEGRATION BY PARTS FORMULA:

$$\int u\,dv = uv - \int v\,du$$

Here are some observations regarding this formula:


The classic problem that requires integration by parts is integrating the natural logarithm function.

Let $\,dv = dx\,$, so that $\,v = x\,$.
Let $u = \ln x\,$, so that $\,du = \frac 1x\,dx\,$.
$\int \overbrace{\ln x}^{u}\ \overbrace{\ dx\ }^{dv} = \overbrace{(\ln x)}^{u}\cdot\overbrace{\ \ x\ \strut\ }^{v} - \int \overbrace{\strut\ \ x\ \ }^{v}\cdot\overbrace{\frac{1}{x}\,dx}^{du} = x\ln x - \int (1)\,dx = x\ln x - x + C$


Let $u = x\,$, so that $\,du = \,dx\,$.
Let $\,dv = {\text{e}}^x\,dx\,$, so that $\,v = {\text{e}}^x\,$.

$\int x{\text{e}}^x\,dx = x{\text{e}}^x - \int {\text{e}}^x\,dx = x{\text{e}}^x - {\text{e}}^x + C $
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