Evaluate the following integrals: e^ 3x 4e^ 6x -9 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{\text{e}^{\text{3x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$

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Answer

Let $\text{I}=\int\frac{\text{e}^{3\text{x}}}{4\text{e}^{6\text{x}}-9}\text{dx}$
Let $\text{e}^{3\text{x}}=\text{t}$
$\Rightarrow3\text{e}^{3\text{x}}\text{ dx = dt}$
$\Rightarrow\text{e}^{3\text{x}} \text{dx}=\frac{\text{dt}}{3}$
$\text{I}=\frac{1}3{}\int\frac{\text{dt}}{4\text{t}^2-9}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\frac{9}{4}}$
$=\frac{1}{12}\int\frac{\text{dt}}{\text{t}^2-\big(\frac{3}{2}\big)^2}$
$=\frac{1}{12}\times\frac{1}{2\big(\frac{3}{2}\big)}\log\Bigg|\frac{\text{t}-\frac{3}{2}}{\text{t}+\frac{3}{2}}\Bigg|+\text{C}$ $\bigg[\text{Since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\bigg]$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{t}-3}{2\text{t}+3}\bigg|+\text{C}$
$\text{I}=\frac{1}{36}\log\bigg|\frac{2\text{e}^{3\text{x}}-3}{2\text{e}^{3\text{x}}+3}\bigg|+\text{C}$

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