Evaluate the following integrals: e^ x e^ 2x +5e^ x +6 dx - Vidyadip

Question

Evaluate the following integrals:
$\int{\frac{\text{e}^{\text{x}}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}}\text{dx}$

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Answer

$\int\frac{\text{e}^\text{x}\text{dx}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}$
Let $\text{e}^{\text{x}}=\text{t}$
$\Rightarrow \text{e}^{\text{x} }\text{dx = dt}$
Now, $\int\frac{\text{e}^{\text{x}}\text{ dx}}{\text{e}^{2\text{x}}+5\text{e}^{\text{x}}+6}$
$=\int\frac{\text{dt}}{\text{t}^2+5\text{t}+6}$
$=\int\frac{\text{dt}}{\text{t}^2+5\text{t}+\big(\frac{5}{2}\big)^2-\big(\frac{5}{2}\big)^2+6}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\frac{25}{4}+6}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\frac{25+24}{4}}$
$=\int\frac{\text{dt}}{\big(\text{t}+\frac{5}{2}\big)^2-\big(\frac{1}{2}\big)^2}$
$=\frac{1}{2\times\frac{1}{2}}\log\Bigg|\frac{\text{t}+\frac{5}{2}-\frac{1}{2}}{\text{t}+\frac{5}{2}+\frac{1}{2}}\Bigg|+\text{C}$
$=\log\bigg|\frac{\text{t}+2}{\text{t}+3}\bigg|+\text{C}$
$=\log\bigg|\frac{\text{e}^\text{x}+2}{\text{e}^\text{x}+3}\bigg|+\text{C}$

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