Question

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

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

Answer 1

The correct option is B $\frac{x^{4m}}{2m(x^{2m}+x^m+1{)}^2}$

$\int \frac{x^{5m-1}+2x^{4m-1}}{(x^{2m}+x^m+1{)}^3}dx$
Rearrange the equation
$\int \frac{2x^{4m-1}(x^{2m}+x^m+1)-x^{4m}(x^{2m-1}+x^{m-1})}{(x^{2m}+x^m+1{)}^3}dx$.............. {1}
As we know that $\int \frac{f^{\prime }\left(x\right)g^2\left(x\right)-2f\left(x\right)g^{\prime }\left(x\right)g\left(x\right)}{g^4\left(x\right)}dx=\frac{f\left(x\right)}{g^2\left(x\right)}$
trying to reduce the equation {1} in this form, we get
$\int \frac{2m\times 4m{x}^{4m-1}(x^{2m}+x^m+1{)}^2-x^{4m}\times 4m^2\times (x^{2m}+x^m+1\left)\right(2m{x}^{2m-1}+m{x}^{m-1})}{(x^{2m}+x^m+1{)}^3}dx$
separate into two quantities and we get,
$=\frac{x^{4m}}{2m(x^{2m}+x^m+1{)}^2}$