Integration by parts

Solution

1. Let $f(x) = \log x$ and $g'(x) = x$. Then $f'(x) = \dfrac{1}{x}$ and $g(x) = \frac{1}{2}x^{2}$. Use integration by parts, $\displaystyle\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx$.

2. $$\begin{aligned}\int x\log x\,dx &= \frac{1}{2}x^{2}\log x - \int \frac{1}{2}x^{2} \cdot \frac{1}{x}\,dx \\ &= \frac{1}{2}x^{2}\log x - \frac{1}{2}\int x\,dx \\ &= \frac{1}{2}x^{2}\log x - \frac{1}{4}x^{2} + C\end{aligned}$$

Solution

1. Take $f(x) = x^2$ and $g'(x) = e^{x}$, and integrate by parts twice. Use integration by parts, $\displaystyle\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx$.

2. $$\begin{aligned}\int x^{2}e^{x}\,dx &= x^{2}e^{x} - 2\int xe^{x}\,dx \\ &= x^{2}e^{x} - 2\left\{xe^{x} - \int e^{x}\,dx\right\} \\ &= x^{2}e^{x} - 2xe^{x} + 2e^{x} + C \\ &= (x^{2} - 2x + 2)e^{x} + C\end{aligned}$$

Solution

1. Let $t = x^2$. Then $dt = 2x\,dx$, so $x^3 e^{x^2}\,dx = x^2 e^{x^2} \cdot x\,dx = \dfrac{1}{2}te^t\,dt$. Then integrate by parts:

   $$\begin{aligned}\int x^{3}e^{x^{2}}\,dx &= \frac{1}{2}\int te^{t}\,dt \\ &= \frac{1}{2}\left(te^{t} - \int e^{t}\,dt\right) \\ &= \frac{1}{2}(t - 1)e^{t} + C \\ &= \frac{1}{2}(x^{2} - 1)e^{x^{2}} + C\end{aligned}$$

Solution

1. Let $f(x) = x$ and $g'(x) = \cos x$. Then $g(x) = \sin x$. Use integration by parts, $\displaystyle\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx$.

2. $$\begin{aligned}\int x\cos x\,dx &= x\sin x - \int \sin x\,dx \\ &= x\sin x + \cos x + C\end{aligned}$$

Solution

1. Write $\log(x + 3) = (x + 3)'\log(x + 3)$ and integrate by parts with $g(x) = x + 3$. Use integration by parts, $\displaystyle\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx$.

2. $$\begin{aligned}\int \log(x + 3)\,dx &= (x + 3)\log(x + 3) - \int (x + 3) \cdot \frac{1}{x + 3}\,dx \\ &= (x + 3)\log(x + 3) - \int dx \\ &= (x + 3)\log(x + 3) - x + C\end{aligned}$$

Taking $g(x) = x$ instead gives $x\log(x + 3) - x + 3\log(x + 3)$, which is the same expression.

Solution

1. Since $(\tan x)' = \dfrac{1}{\cos^2 x}$, integrate by parts with $f(x) = x$ and $g(x) = \tan x$. Also, $\displaystyle\int \tan x\,dx = -\log|\cos x| + C$.

   $$\begin{aligned}\int \frac{x}{\cos^{2} x}\,dx &= x\tan x - \int \tan x\,dx \\ &= (x\tan x + \log|\cos x|) + C\end{aligned}$$