Question

$\int\limits \frac{x^3 -1}{x^3 + x} dx = $

• A. $x - \log x + \log \left(x^{2} + 1\right)- \tan^{-1} x + c$
• B. $x - \log x + \frac{1}{2} \log \left(x^{2} + 1\right)- \tan^{-1} x + c$
• C. $x + \log x + \log \left(x^{2} + 1\right)- \tan^{-1} x + c$
• D. $x + \log x + \frac{1}{2} \log \left(x^{2} + 1\right)- \tan^{-1} x + c$

Answer 1

Answer: (B)

Let $I = \int\limits \frac{x^3 -1}{x^3 + x} dx = \int\limits \left(1 - \frac{x + 1}{x^3 + x}\right) dx $
$= \int\limits 1dx - \int\limits \frac{x + 1}{x(x^2 +1 )} dx = x - \int\limits \frac{x + 1}{x(x^2 +1 )}$ ....(i)
Now $ \frac{x + 1}{x(x^2 +1 )} = \frac{A}{x} + \frac{Bx + C}{x^2 +1}$
(By using partial fractions)
$\Rightarrow \:\: x + 1 = A(x^2 + 1) +(Bx+ C)x$
$\Rightarrow \:\: x + 1 = (A + B)x^2 + Cx + A$
Comparing coefficients of $x^2, x$ and constant,
we get
$A + B = 0, C = 1, A = 1 $
$ \Rightarrow \: B = -1$
$\therefore \:\: $ From (i), we get
$ I = x - \int\limits \frac{1}{x} dx - \int\limits \frac{1 -x }{x^2 + 1} dx $
$ = x - \log \, x - \int\limits \frac{1}{x^2 + 1} dx + \frac{1}{2} \int\limits \frac{2x}{x^2 + 1} dx $
$ = x - \log \, x - \tan^{-1} x + \frac{1}{2} \log (x^2 + 1) + c$