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 $