Question

Primitive of $\frac{3x^4-1}{{\left(x^4+x+1\right)}^2}$ w.r.t. $x$ is :

• A. $\frac{x}{x^4+x+1}+c$
• B. $-\frac{x}{x^4+x+1}+c$
• C. $\frac{x+1}{x^4+x+1}+c$
• D. $-\frac{x+1}{x^4+x+1}+c$

Answer 1

Answer: (B)

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