Question

Solve:

$\int \frac{1}{(x+1)(x^2+1)}dx$

Answer 1

Given the integral,

$\int \frac{1}{(x+1)(x^2+1)}dx$

using partial fraction we get,

$\int \frac{1}{(x+1)(x^2+1)}dx=\int [\frac{1}{2(x+1)}-\frac{x-1}{2(x^2+1)}]dx=\frac{1}{2}\int \frac{1}{x+1}dx-\frac{1}{2}\int \frac{x-1}{x^2+1}dx$

Now, for $\int \frac{1}{x+1}dx$

Let, $u=x+1\Rightarrow \frac{du}{dx}=1\Rightarrow du=dx$

Substituting the values of $u$ and $du$ we get,

$\int \frac{1}{x+1}dx=\int \frac{1}{u}du=\ln \left(u\right)=\ln (x+1)$

for $\int \frac{x-1}{x^2+1}dx$

Expanding the integral,

$\int \frac{x-1}{x^2+1}dx=\int \frac{x}{x^2+1}dx-\int \frac{1}{x^2+1}dx$

Again, for $\int \frac{x}{x^2+1}dx$

Let,

$u=x^2+1\Rightarrow \frac{du}{dx}=2x\Rightarrow dx=\frac{1}{2x}du\therefore \int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{1}{u}du=\frac{\ln \left(u\right)}{2}=\frac{\ln (x^2+1)}{2}$

And

$\int \frac{1}{x^2+1}dx=\arctan \left(x\right)\therefore \int \frac{x}{x^2+1}dx-\int \frac{1}{x^2+1}dx=\frac{\ln (x^2+1)}{2}-\arctan \left(x\right)$

So,

$\frac{1}{2}\int \frac{1}{x+1}dx-\frac{1}{2}\int \frac{x-1}{x^2+1}dx=\frac{\ln (x+1)}{2}-\frac{\ln (x^2+1)}{4}+\frac{\arctan \left(x\right)}{2}\therefore \int \frac{1}{(x+1)(x^2+1)}dx=\frac{\ln (x+1)}{2}-\frac{\ln (x^2+1)}{4}+\frac{\arctan \left(x\right)}{2}+C.$