# Evaluate : $\int \large\frac{x^4dx}{(x-1)(x^2+1)}$

Toolbox:
• $\int \large\frac{dx}{x^2+1}$$=\tan^{-1}x+c Step 1: Let I=\int \large\frac{x^4dx}{(x-1)(x^2+1)} \large\frac{x^4}{(x-1)(x^2+1)}=\frac{x^4}{x^3+x-x^2-1} This is an improper rational function \Rightarrow (x+1)-\large\frac{1}{(x^3-x^2+x-1)} \Rightarrow x+1-\large\frac{1}{(x-1)(x^2+1)} Step 2: Consider \large\frac{1}{(x-1)(x^2+1)} This can be resolved into partial fractions as \large\frac{A}{(x-1)}+\frac{Bx+C}{(x^2+1)} 1=A(x^2+1)+(x-1)(Bx+C) Put x=1 1=2A\Rightarrow A=\large\frac{1}{2} Comparing the coefficient of x^2 term 0=A+B\Rightarrow B=-\large\frac{1}{2} Comparing the coefficient of constant term 1=A-C C=-\large\frac{1}{2} Step 3: \therefore I=\int (x+1)dx-\int\large\frac{dx}{2(x-1)}+\int \frac{-1/2x-1/2}{x^2+1}$$dx-\large\frac{1}{2}\int \large\frac{dx}{x^2+1}$
On integrating we get,
$I=\large\frac{x^2}{2}$$+x-\large\frac{1}{2}$$\log(x-1)-\large\frac{1}{2}\times \frac{1}{2}$$\log (x^2+1)-\large\frac{1}{2}$$\tan^{-1}(x)+C$
$\;\;=\large\frac{x^2}{2}$$+x-\large\frac{1}{2}$$\log(x-1)-\large\frac{1}{4}$$\log(x^2+1)-\large\frac{1}{2}$$\tan^{-1}x+C$
$\;\;=\large\frac{x^2}{2}$$+x-\large\frac{1}{4}\frac{\log(x-1)^2}{x^2+1}-\large\frac{1}{2}$$\tan^{-1}x+C$