Question

Primitive of $\frac{3x^4-1}{(x^4+x+1{)}^2}$ 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

The correct option is B $-\frac{x}{(x^4+x+1)}+c$

$\int \frac{{3x}^4-1}{{(x^4+x+1)}^2}dx=\int \frac{{3x}^2-\frac{1}{x^2}}{{(x^3+1+\frac{1}{x})}^2}dx$
Substitute $x^3+1+\frac{1}{x}=t\Rightarrow ({3x}^2-\frac{1}{x^2})dx=dt$
$\therefore \int \frac{dt}{t^2}=-\frac{1}{t}+c=-\frac{1}{x^3+1+\frac{1}{x}}+c$
$=-\frac{x}{x^4+x+1}+c$