Evaluate the following integrals: 1 4x^2+12x+5 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{4\text{x}^2+12\text{x}+5}\text{dx}$
$=\frac{1}{4}\int\frac{1}{\text{x}^2+3\text{x}+\frac{5}{4}}\text{dx}$
$=\frac{1}4{}\int\frac{1}{\text{x}^2+2\times\text{x}\times\big(\frac{3}{2}\big)+\big(\frac{3}{2}\big)^2-\big(\frac{3}{2}\big)^2+\frac{5}{4}}\text{dx}$
$\text{I}=\frac{1}{4}\int\frac{1}{\Big(\text{x}+\frac{3}{2}\Big)^2-1}\text{dx}$
Let $\Big(\text{x}+\frac{3}{2}\Big)=\text{t}\ \dots(1)$
$\Rightarrow\text{dx = dt}$
So,
$\text{I}=\frac{1}{4}\int\frac{1}{\text{t}^2-(1)^2}\text{dt}$
$\text{I}=\frac{1}{4}\times\frac{1}{2\times(1)}\log\bigg|\frac{\text{t}-1}{\text{t}+1}\bigg|+\text{C}$ $\Big[\text{Since,} \int\frac{1}{\text{x}^2-\text{a}^2}\text{dx}=\frac{1}{2\text{a}}\log\bigg|\frac{\text{x}-\text{a}}{\text{x+a}}\bigg|+\text{C}\Big]$
$\text{I}=\frac{1}{8}\log\Bigg|\frac{\text{x}+\frac{3}{2}-1}{\text{x}+\frac{3}{2}+1}\Bigg|+\text{C}$ [using (1)]
$\text{I}=\frac{1}{8}\log\bigg|\frac{2\text{x}+1}{2\text{x}+5}\bigg|+\text{C}$

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