$\int e^{3 l o g x} (x^{4} + 1)^{- 1} d x .$

Open in App

Solution

$L e t I = \int e^{3 l o g x} (x^{4} + 1)^{- 1} d x = \int \frac{e^{l o g x^{3}}}{(x^{4} + 1)} d x = \int \frac{x^{3}}{(x^{4} + 1)} d x (∵ e^{l o g x} = x) P u t x^{4} + 1 = t \Rightarrow 4 x^{3} d x = d t \Rightarrow d x = \frac{d t}{4 x^{3}} ∴ I = \int \frac{x^{3}}{t} \frac{d t}{4 x^{3}} = \frac{1}{4} \int \frac{1}{t} d t = \frac{1}{4} l o g | t | + C = \frac{1}{4} l o g | x^{4} + 1 | + C$

Suggest Corrections

Similar questions

e^{3 log x} (x^{4} + 1)^{- 1}

\int \frac{1}{x^{4} - 1} d x

\int \frac{1}{x^{4} - 1} d x

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

\int \frac{1}{x^{13} (x^{6} - 1)} d x

Join BYJU'S Learning Program