Question

If m is a non - zero number and $\int \frac{x^{5m-1}+2x^{4m-1}}{{\left(x^{2m}+x^m+1\right)}^3}dx=f\left(x\right)+c,$
then f(x) is :

• A. $\frac{x^{5m}}{2m{\left(x^{2m}+x^m+1\right)}^2}$
• B. $\frac{x^{4m}}{2m{\left(x^{2m}+x^m+1\right)}^2}$
• C. $\frac{2m\left(x^{5m}+x^{4m}\right)}{{\left(x^{2m}+x^m+1\right)}^2}$
• D. $\frac{\left(x^{5m}-x^{4m}\right)}{2m{\left(x^{2m}+x^m+1\right)}^2}$

Answer 1

Answer: (B)

$\int \frac{x^{5m-1}+2x^{4m-1}}{{\left(x^{2m}+x^m+1\right)}^3}dx$
$=\int \frac{x^{5m-1}+2x^{4m-1}}{x^{6m}{\left(1+x^{-m}+x^{-2m}\right)}^3}dx$
$=\int \frac{x^{-m-1}+2x^{-2m-1}}{{\left(1+x^{-m}+x^{-2m}\right)}^3}dx$
Put $t=1+x^{-m}+x^{-2m}$
$\therefore \frac{dt}{dx}=-m{x}^{-m-1}-2m{x}^{-2m-1}$
$\Rightarrow \frac{dt}{-m}=\left(x^{-m-1}+2x^{-2m-1}\right)dx$
$\therefore \int \frac{x^{5m-1}+2x^{4m-1}}{{\left(x^{2m}+x^m+1\right)}^3}dx$
$=\frac{1}{-m}\int t^{-3}dt=\frac{1}{2m{t}^2}+C$
$=\frac{1}{2m\left(1+x^{-m}+x^{-2m}\right)}+C$
$=\frac{x^{4m}}{2m{\left(x^{2m}+x^m+1\right)}^2}+C$
$\therefore f\left(x\right)=\frac{x^{4m}}{2m{\left(x^{2m}+x^m+1\right)}^2}$