Let $I = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x, J = \frac{e^{3 x}}{e^{4 x} + e^{2 x} + 1}$ then the
value of $I - J$ equals

Moderate

Solution

$I - J = \int \frac{e^{x}}{e^{4 x} + e^{2 x} + 1} d x - \int \frac{e^{3 x}}{e^{4 x} + e^{2 x} + 1} d x$
$= \int \frac{e^{x} (1 - e^{2 x})}{e^{4 x} + e^{2 x} + 1} d x$
Put $e^{x} = t, e^{x} d x = d t$
$\begin{array}{l} I - J = \int \frac{(1 - t^{2})}{t^{4} + t^{2} + 1} d t = \int \frac{\frac{1}{t^{2}} - 1}{t^{2} + \frac{1}{t^{2}} + 1} d t \\ = - \int \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 1} = \int \frac{d u}{1 - u^{2}} \\ u = t + \frac{1}{t} \end{array}$
$\begin{aligned} = \frac{1}{2} \log |\frac{1 + u}{1 - u}| + C \\ = \frac{1}{2} \log |\frac{e^{x} + e^{- x} + 1}{e^{x} + e^{- x} - 1}| + C \\ = \frac{1}{2} \log |\frac{e^{2 x} + e^{x} + 1}{e^{2 x} - e^{x} + 1}| + C . \end{aligned}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

SCAN ME