Question

Find $\int \frac{x^4+1}{x(x^2+1{)}^2}dx.$

Answer 1

Let $I=\int \frac{x^4+1}{x(x^2+1{)}^2}dx=\int \frac{x^4+1+2x^2-2x^2}{x(x^2+1{)}^2}dx=\int \frac{(x^2+1{)}^2-2x^2}{x(x^2+1{)}^2}$
$\Rightarrow I=\int \frac{(x^2+1{)}^2}{x(x^2+1{)}^2}dx-\int \frac{2x^2}{x(x^2+1{)}^2}dx\Rightarrow I=\int \frac{1}{x}dx-\int \frac{2x}{(x^2+1{)}^2}dx$
In second integral, put $x^2+1=y\Rightarrow 2xdx=dy\therefore I=log|x|-\int \frac{dy}{y^2}=log|x|+\frac{1}{y}+C$
Thus, $I=log|x|+\frac{1}{x^2+1}+C.$
Alternative: $I=\int \frac{x^4+1}{x(x^2+1{)}^2}dx=\int \frac{(x^4+1)x}{x^2(x^2+1{)}^2}dx$ [Put $x^2=t\Rightarrow xdx=\frac{dt}{2}]$
$\Rightarrow I=\frac{1}{2}\int \frac{t^2+1}{t(t+1{)}^2}dt$
Consider $\frac{t^2+1}{t(t+1{)}^2}=\frac{A}{t}+\frac{B}{t+1}+\frac{C}{(t+1{)}^2}\Rightarrow t^2+1=A(t+1{)}^2+Bt(t+1)+Ct$
On comparing the coefficients of like terms both sides, we get: A=1, B=0, C=-2
$\therefore I=\frac{1}{2}\int \{\frac{1}{t}+\frac{0}{t+1}-\frac{2}{(t+1{)}^2}\}dt=\frac{1}{2}\{log|t|+\frac{2}{t+1}\}+C=\frac{1}{2}\{log|x^2|+\frac{2}{x^2+1}\}+C$
Therefore,$I=log|x|+\frac{1}{x^2+1}+C.$