$\int 16 x {sec}^{2} (2 x^{2} + 1) d x$

Open in App

Solution

$\int 16 x {sec}^{2} (2 x^{2} + 1) d x$
Let $2 x^{2} + 1 = t ⟹ 4 x d x = d t \int 4 s e c^{2} t d t 4 tan t 4 tan (2 x^{2} + 1) + c$

Suggest Corrections

Similar questions

\int \frac{x^{3} + x + 1}{x^{2} - 1} d x

\int \frac{x^{3} + x + 1}{x^{2} - 1} d x

\int \frac{x^{4} + x^{2} + 1}{x^{2} - x + 1} d x =

\int \frac{x^{5}}{x^{2} + 1} d x =

\frac{\int_{0}^{π / 2} {(sin x)}^{\sqrt{2} + 1} d x}{\int_{0}^{π / 2} {(sin x)}^{\sqrt{2} - 1} d x}

I_{n} = \int_{0}^{\infty} e^{- x} x^{n - 1} d x

\int_{0}^{\infty} e^{- λ x} x^{n - 1} d x

\int \sqrt{e^{2 x} - 1} d x

Join BYJU'S Learning Program