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

$\frac{x^2}{2}+x+\frac{1}{2}\log\left|\frac{x-1}{\sqrt{x^2+1}}\right|-\frac{1}{2}\tan^{-1}x+c$

The correct answer is Option (3) → $\frac{x^2}{2}+x+\frac{1}{2}\log\left|\frac{x-1}{\sqrt{x^2+1}}\right|-\frac{1}{2}\tan^{-1}x+c$

$\int \frac{x^4}{(x-1)(x^2+1)} dx = \int \frac{x^4}{x^3-x^2+x-1} dx$

Using long division:

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

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

For partial fractions:

$\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

$1 = A(x^2+1) + (Bx+C)(x-1)$

$1 = (A+B)x^2 + (C-B)x + (A-C)$

Comparing coefficients:

$A+B = 0 ⇒A = -B$

$C-B = 0 ⇒C = B$

$A-C = 1$

$-2C-=1 ⇒C = -\frac{1}{2}$

$∴B = -\frac{1}{2}$ and $A = \frac{1}{2}$

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

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

$=\frac{1}{2}\left[\log|x-1|-\frac{1}{2}\log|x^2+1|-\tan^{-1}x\right]+c$

$=\frac{1}{2}\log\left|\frac{x-1}{\sqrt{x^2+1}}\right|-\frac{1}{2}\tan^{-1}x+c$

$∴\int \frac{x^4}{(x-1)(x^2+1)} dx =\frac{x^2}{2}+x+\frac{1}{2}\log\left|\frac{x-1}{\sqrt{x^2+1}}\right|-\frac{1}{2}\tan^{-1}x+c$