The integral $\int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x$ is equal to: (where C is a constant of integration)

\frac{x^{4}}{(2 x^{4} + 3 x^{2} + 1)^{3}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x^{12}}{6 (2 x^{4} + 3 x^{2} + 1)^{3}} + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{x^{4}}{6 (2 x^{4} + 3 x^{2} + 1)^{3}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x^{12}}{(2 x^{4} + 3 x^{2} + 1)^{3}} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{x^{12}}{6 (2 x^{4} + 3 x^{2} + 1)^{3}} + C$
$\int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x$

$\int \frac{⟮ \frac{3}{x^{3}} + \frac{2}{x^{5}} ⟯ d x}{⟮ 2 + \frac{3}{x^{2}} + \frac{1}{x^{4}} ⟯^{4}}$

Let $⟮ 2 + \frac{3}{x^{2}} + \frac{1}{x^{4}} ⟯ = t$

$- \frac{1}{2} \int \frac{d t}{t^{4}} = \frac{1}{6 t^{3}} + C \Rightarrow \frac{x^{12}}{6 (2 x^{4} + 3 x^{2} + 1)^{4}} + C$

Suggest Corrections

Similar questions

\int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x

C

\int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x

C

\int \frac{3 x^{13} + 2 x^{11}}{(2 x^{4} + 3 x^{2} + 1)^{4}} d x

C

y = (x^{2} + 1) (x^{3} + 3)

\int (x^{6} + x^{4} + x^{2}) \sqrt{2 x^{4} + 3 x^{2} + 6} d x = \frac{(α x^{6} + β x^{4} + γ x^{2})^{3 / 2}}{18} + C

(β + γ - α) .

Join BYJU'S Learning Program