Question

Solve $\int \frac{x^4}{x^4-1}dx$

• A. $x+\frac{1}{6}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c$
  
• B. $x+\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c$
• C. $=x+\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c$
  
• D. $=x+\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|+\frac{1}{2}{\tan }^{-1}+c$
  

Answer 1

The correct option is B $x+\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c$

$\int \frac{x^4}{x^4-1}dx$

$\int \frac{x^4-1}{x^4-1}+\frac{1}{x^4-1}dx$

$\int dx+\int \frac{1}{x^4-1}dx$

$=x+\int \frac{1}{x^4-1}dx$

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

$=x+\frac{1}{2}\int \frac{1}{(x^2-1)}dx-\frac{1}{2}\int \frac{1}{(x^2+1)}dx$ $[\therefore \int \frac{1}{x^2-a^2}dx=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+c]$

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

$=x+\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c$

Hence, the answer is $x+\frac{1}{4}\log \left|\frac{x-1}{x+1}\right|-\frac{1}{2}{\tan }^{-1}+c.$