The integral is equal towhere C is an arbitrary constant.

Engineering

Mathematics

Reduction Formulae in Indefinite Integration

Question

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

where C is an arbitrary constant.

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

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

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

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

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

$\int \frac{(2 x^{12} + 5 x^{9})}{x^{15} (1 + \frac{1}{x^{2}} + \frac{1}{x^{5}})} dx = \int \frac{\frac{2}{x^{3}} + \frac{5}{x^{6}}}{{(1 + \frac{1}{x^{2}} + \frac{1}{x^{5}})}^{3}} dx$

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

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

$= \int \frac{- dt}{t^{3}} = \frac{1}{2 t^{2}} + C = \frac{x^{10}}{2 {(x^{5} + x^{3} + 1)}^{2}} + C$