Question

$\int \frac{2x^{12}+5x^9}{(x^5+x^3+1{)}^3}dx$ is equal to

• A. $\frac{x^{10}}{(x^5+x^3+1{)}^3}+C$
• B. $\frac{x^5}{(x^5+x^3+1{)}^2}+C$
• C. $\frac{x^{10}}{2(x^5+x^3+1{)}^2}+C$
• D. $\frac{-x^5}{(x^5+x^3+1{)}^2}+C$

Answer 1

The correct option is C $\frac{x^{10}}{2(x^5+x^3+1{)}^2}+C$

$\int \frac{2x^{12}+5x^9}{(x^5+x^3+1{)}^3}dx$

Divide numerator & denominator by highest power of expression i.e., '$x^{15}$'

$\int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{{(1+\frac{1}{x^2}+\frac{1}{x^5})}^3}dx$

Put $1+\frac{1}{x^2}+\frac{1}{x^5}=t$

$\Rightarrow \frac{-2}{x^3}-\frac{5}{x^6}dx=dt$ or $\frac{2}{x^3}+\frac{5}{x^6}dx=dt$

$\Rightarrow \frac{\frac{2}{x^3}+\frac{5}{x^6}}{{(1+\frac{1}{x^2}+\frac{1}{x^5})}^3}dx=-\int \frac{1}{t^3}dt$

$=-\frac{t^{-2}}{-2}+C$

$=\frac{1}{2t^2}+C$

$=\frac{1}{2{(1+\frac{1}{x^2}+\frac{1}{x^5})}^2}+C$

$=\frac{x^{10}}{2(x^5+x^3+1{)}^2}+C$.