7.2 | Integration by Parts


Integration by parts formula is given by

$$
\int f(x) g'(x) \; dx = f(x) g(x) – \int f'(x) g(x) \; dx
$$
$$
\int f(x) g'(x) \; dx = f(x) g(x) – \int f'(x) g(x) \; dx
$$
$$
\begin{aligned}
\int & f(x) g'(x) \; dx \\
& = f(x) g(x) – \int f'(x) g(x) \; dx
\end{aligned}
$$
Find
$$
\int x^3 e^x \; dx
$$


$$
\begin{array}{r c l}
x^3 & _{\displaystyle \overset{+1}{\searrow}} & e^x \\
3x^2 &_{\displaystyle \overset{-1}{\searrow}} & e^x \\
6x & _{\displaystyle \overset{+1}{\searrow}} & e^x \\
6 & _{\displaystyle \overset{-1}{\searrow}} & e^x \\
0 & \overset{+1}{\rightarrow} & e^x
\end{array}
$$

By applying integration by parts 4 times, we obtain:

$$
\begin{aligned}
\int x^3 e^x \; dx & = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C
\end{aligned}
$$
$$
\begin{aligned}
\int x^3 e^x \; dx & = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C
\end{aligned}
$$
$$
\begin{aligned}
\int & x^3 e^x \; dx \\
& = x^3 e^x -3x^2 e^x + 6xe^x – 6e^x + C
\end{aligned}
$$
Find
$$
\int t \ln t \; dt
$$


$$
\begin{array}{r c l}
\ln t & _{\displaystyle \overset{+1}{\searrow}} & t \\
\frac{1}{t} & \overset{-1}{\rightarrow} & \frac{t^2}{2} \\
\end{array}
$$

By applying integration by parts once, we obtain:
$$
\begin{aligned}
\int t \ln t \; dt & = t \ln|t| + \int \frac{-t}{2} \; dt \\
& = t \ln|t| – \int \frac{t}{2} \; dt \\
& = t \ln|t| – \frac{t^2}{4} + C \\
\end{aligned}
$$

Find
$$
\int e^{\theta} \sin \theta \; d \theta
$$


$$
\begin{array}{r c l}
e^{\theta} & _{\displaystyle \overset{+1}{\searrow}} & \sin \theta \\
e^{\theta} & _{\displaystyle \overset{-1}{\searrow}} & – \cos \theta \\
e^{\theta} & \overset{+1}{\rightarrow} & – \sin \theta \\
\end{array}
$$

By applying integration by parts twice, we obtain:

$$
\begin{aligned}
\int e^{\theta} \sin \theta \; d \theta & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\
& = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\
\end{aligned}
$$
$$
\begin{aligned}
\int e^{\theta} \sin \theta \; d \theta & = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\
& = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\
\end{aligned}
$$
$$
\begin{aligned}
& \int e^{\theta} \sin \theta \; d \theta \\
& = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta + \int -e^{\theta} \sin \theta \; d\theta \\
& = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta – \int e^{\theta} \sin \theta \; d \theta \\
\end{aligned}
$$

So,
$$
2 \int e^{\theta} \sin \theta \; d \theta = -e^{\theta} \cos \theta+ e^{\theta} \sin \theta
$$
and
$$
\int e^{\theta} \sin \theta \; d \theta = \frac{-e^{\theta} \cos \theta+ e^{\theta} \sin \theta}{2} + C
$$

Find
$$
\int (3t^2-5t+10) e^{2t} \; dt
$$


$$
\begin{array}{r c l}
3t^2-5t+10 & _{\displaystyle \overset{+1}{\searrow}} & e^{2t} \\
6t-5 &_{\displaystyle \overset{-1}{\searrow}} & \frac{e^{2t}}{2} \\
6 & _{\displaystyle \overset{+1}{\searrow}} & \frac{e^{2t}}{4} \\
0 & _{\displaystyle \overset{-1}{\rightarrow}} & \frac{e^{2t}}{8} \\
\end{array}
$$

By applying integration by parts 3 times, we obtain:

$$
\begin{aligned}
\int (3t^2-5t+10) e^{2t} \; dt & = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} + \frac{6}{8}e^{2t} + C\\
& = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} + \frac{3}{4}e^{2t} + C
\end{aligned}
$$
$$
\begin{aligned}
& \int (3t^2-5t+10) e^{2t} \; dt \\
& = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} \\
& \; \; + \frac{6}{8}e^{2t} + C\\
& = e^{2t} \left( \frac{3t^2-5t+10}{2} -\frac{6t-5}{4} + \frac{3}{4}\right) + C
\end{aligned}
$$
$$
\begin{aligned}
& \int (3t^2-5t+10) e^{2t} \; dt \\
& = \frac{3t^2-5t+10}{2} e^{2t} -\frac{6t-5}{4} e^{2t} \\
& \; \; + \frac{6}{8}e^{2t} + C\\
& = e^{2t} \left( \frac{3t^2-5t+10}{2} -\frac{6t-5}{4} + \frac{3}{4}\right) + C
\end{aligned}
$$

E 7.2 Exercises