Question

Solve the following:
(a) $\int \frac{x^4+x^2+1}{x^2-x+1}dx=$ ?
(b) $\int \frac{x^5}{x^2+1}dx=$ ?

Answer 1

$a)$ Given the integral,

$\int \frac{x^4+x^2+1}{x^2-x+1}dx=\int \frac{(x^2-x+1)(x^2+x+1)}{(x^2-x+1)}dx=\int (x^2+x+1)dx=\int \left(x^2\right)dx+\int \left(x\right)dx+\int \left(1\right)dx=\frac{x^3}{3}+\frac{x^2}{2}+x+C=\frac{2x^3+3x^2+6x}{6}+C.$

$b)$ Given the integral,

$\int \frac{x^5}{x^2+1}dx$

Let,

$u=x^2+1\Rightarrow \frac{du}{dx}=2x\Rightarrow du=2xdx$

and

$x^2=u-1\Rightarrow x^4={(u-1)}^2$

Substituting these values we get,

$\int \frac{x^5}{x^2+1}dx=\int \frac{x\cdot x^4}{x^2+1}dx=\frac{1}{2}\int \frac{x^{4}2x}{x^2+1}dx=\frac{1}{2}\int \frac{{(u-1)}^2}{u}du$

For,

$\int \frac{{(u-1)}^2}{u}du=\int \left[\frac{u^2-2u+1}{u}\right]du=\int [u-2+\frac{1}{u}]du=\int \left(u\right)du-\int \left(2\right)du+\int \left(\frac{1}{u}\right)du=\frac{u^2}{2}-2u+\ln \left(u\right)=\frac{{(x^2+1)}^2}{2}-2(x^2+1)+\ln (x^2+1)\therefore \frac{1}{2}\int \frac{{(u-1)}^2}{u}du=\frac{{(x^2+1)}^2}{4}+\frac{\ln (x^2+1)}{2}-(x^2+1)$

Hence,

$\int \frac{x^5}{x^2+1}dx=\frac{{(x^2+1)}^2}{4}+\frac{\ln (x^2+1)}{2}-(x^2+1)+C.$