Question

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

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

Answer 1

Answer: (B)

$\int \frac{2x^{12}+5x^9}{{\left[x^5\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)\right]}^3}=\int \frac{2x^{12}+5x^9}{x^{15}{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)}^3}dx$
Dividing numerator and denominator by $x^{15}$ we get,
$=\int \frac{\frac{2}{x^3}+\frac{5}{x^6}}{{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)}^3}dx$
Put $\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)=t$
$\frac{-2}{x^3}-\frac{-5}{x^6}=\frac{dt}{dx}$
$\left(\frac{2}{x^3}+\frac{5}{x^6}\right)dx=-dt$
$=\int \frac{-dt}{t^3}$
$=\frac{-t^{-3+1}}{-3+1}+C=\frac{1}{2}\times \frac{1}{t^2}+C$
$=\frac{1}{2}\frac{1}{{\left(1+\frac{1}{x^2}+\frac{1}{x^5}\right)}^2}+C$
$=\frac{1}{2}\frac{x^{10}}{{\left(x^5+x^3+1\right)}^2}+C$