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

(A) $\frac{\mathrm{x}^{10}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}$

(B) $\frac{\mathrm{x}^{5}}{2\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}$

(C) $\frac{-x^{10}}{2\left(x^{5}+x^{3}+1\right)^{2}}+C$

(D) $\frac{-\mathrm{x}^{5}}{\left(\mathrm{x}^{5}+\mathrm{x}^{3}+1\right)^{2}}+\mathrm{C}$

JEE Main · 2016 · 03 Apr 2016 · Offline · Indefinite Integrals

The integral

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

is equal to:

Answer: (A)

Divide numerator and denominator by

x^{15}

, so the integral reads

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

. Put

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

, whose derivative

d t = (- \frac{2}{x ^{3}} - \frac{5}{x ^{6}}) d x

is exactly the negative of the numerator. The integral becomes

- \int \frac{d t}{t ^{3}} = \frac{1}{2 t ^{2}} + C

. Restoring

t

and multiplying through by

x^{10}

gives

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

JEE Main · 2016 · 03 Apr 2016 · Offline · Single correct option