Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{e^{3 x}}{4 e^{6 x}-9} d x$

Solution:

let $I=\int \frac{e^{2 x}}{4 e^{6 x_{-9}}} d x$

Let $\mathrm{e}^{3 \mathrm{x}}=\mathrm{t} \ldots \ldots(\mathrm{i})$

$\Rightarrow 3 e^{3 x} d x=d t$

$I=\frac{1}{3} \int \frac{1}{4 t^{2}-9} d t$

$=\frac{1}{12} \int \frac{1}{t^{2}-\frac{9}{4}} d t$

$I=\frac{1}{12} \int \frac{1}{t^{2}-\left(\frac{3}{2}\right)^{2}} d t$

$I=\frac{1}{36} \log \left|\frac{t-\frac{3}{2}}{t+\frac{3}{2}}\right|+c$

[since, $\left.\int \frac{1}{x^{2}-(a)^{2}} d x=\frac{1}{2 \times a} \log \left|\frac{x-a}{x+a}\right|+c\right]$

$I=\log \left|\frac{2 t-3}{2 t+3}\right|+c$

$I=\log \left|\frac{2 e^{2 x}-3}{2 e^{2 x}+3}\right|+c$ [using (i)]

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