Section 9.2: Integrations by Parts
A Test Problem

Perform the following indefinite integration.

$$\int x \sin(x) dx$$

Answer

$$\int x \sin(x) dx = \sin(x) - x \cos(x) + C$$

Product Rule Recalled

$$(f \cdot g)'(x) = f'(x)g(x) + f(x) g'(x)$$

Inverting the Chain Rule: Integration by Substitution

$$\int f(x) g'(x) dx = f(x)g(x) - \int g(x) f'(x) dx$$

Formalism of Integration by Parts

Often, one finds two functions $u$ and $v$ so that the integrand may be written as $u dv$ where $dv = v'(x) dx$. If we succeed in so doing, then from the equality

$$u(x) v'(x) = u(x) v(x) - v(x) u'(x)$$

we see that if $f(x) = u(x) v'(x)$, then

$$\int f(x) dx = \int u(x) v'(x) dx = \int u dv = u v - \int v du$$

If $\int v du$ is easier to evaluate than is $\int f(x) dx$, then the method succeeds.

An integral revisited

Take $u = x$ and $v = -\cos(x)$ so that $dv = \sin(x) dx$ and $du = dx$. Then,

$$\begin{eqnarray*} \int x \sin(x) dx & = & \int u dv \\ & = & uv - \int v du \\... ...-x \cos(x) + \int \cos(x) dx \\ & = & -x \cos(x) + \sin(x) + C \end{eqnarray*}$$

Example

Integrate:

$$\int x e^x dx$$

Solution

Take $u = x$ and $v = e^x$ so that $du = dx$ and $dv = e^x dx$.

$$\begin{eqnarray*} \int x e^x dx & = & \int u dv \\ & = & uv - \int v du \\ & = & x e^x - \int e^x dx \\ & = & x e^x - e^x + C \end{eqnarray*}$$

Large Another example

Integrate

$$\int e^x \cos(x) dx$$

Solution

Set $u = e^x$ and $v = \sin(x)$ so that $du = e^x$ and $dv = \cos(x) dx$.

Then

$$\begin{eqnarray*} \int e^x \cos(x) dx & = & \int u dv \\ & = & uv - \int v du ... ...t e^x \sin(x) dx \\ & = & e^x \sin(x) + \int (-e^x\sin(x)) dx \end{eqnarray*}$$

Solution, continued

Set $w = e^x$ and $y = \cos(x)$ so that $dw = e^x dx$ and $dy = -\sin(x) dx$.

Then

$$\begin{eqnarray*} \int -e^x \sin(x) dx & = & \int w dy \\ & = & wy - \int y dw \\ & = & e^x \cos(x) - \int e^x \cos(x) dx \end{eqnarray*}$$

Substituting, we have

$$\int e^x \cos(x) dx = e^x \sin(x) + e^x \cos(x) - \int e^x \cos(x) dx$$

Adding $\int e^x \cos(x) dx$ and dividing by $2$ gives

$$\int e^x \cos(x) dx = \frac{e^x}{2}(\sin(x) + \cos(x)) + C$$

A third example

Integrate

$$\int \ln(x) dx$$

Solution

Set $u = \ln(x)$ and $v = x$ so that $du = \frac{dx}{x}$ and $dv = dx$.

$$\begin{eqnarray*} \int \ln(x) dx & = & \int u dv \\ & = & uv - \int v du \\ ... ...x} dx \\ & = & x \ln(x) - \int dx \\ & = & x \ln(x) - x + C \end{eqnarray*}$$

A final example

Integrate

$$\int \frac{x e^{2x}}{x^2 + 4x + 4} dx$$

Solution

Note that $\frac{1}{4x^2 + 4x + 1} = (2x + 1)^{-2}$. Via the substitution $w = 2x+1$ (with $dw = 2 dx$) we find that

$$\begin{eqnarray*} \int \frac{dx}{4x^2 + 4x + 1} & = & \int (2x+2)^{-2} dx \\ &... ...\\ & = & \frac{-1}{2}w^{-1} + C \\ & = & \frac{-1}{4x+2} + C \end{eqnarray*}$$

Solution, continued

Set $u = x e^{2x}$ and $v = \frac{-1}{4x+2}$ (so that $dv = \frac{dx}{x^2 + 4x + 4}$ and $du = e^{2x} + 2x e^{2x} = (2x + 1) e^{2x}$).

$$\begin{eqnarray*} \int \frac{x e^{2x}}{4 x^2 + 4x + 1} dx & = & \int u dv \\ &... ...dx \\ & = & \frac{-x e^{2x}}{4x + 2} + \frac{1}{4} e^{2x} + C \end{eqnarray*}$$