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Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{4 x^{2}+12 x+5} d x$

Solution:

let $I=\int \frac{1}{4 x^{2}+12 x+5} d x$

$=\frac{1}{4} \int \frac{1}{x^{2}+3 x+\frac{5}{4}} d x$

$=\frac{1}{4} \int \frac{1}{x^{2}+2 x \times \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+\frac{5}{4}} d x$

$=\frac{1}{4} \int \frac{1}{\left(x+\frac{3}{2}\right)^{2}-1} d x$

Let $\left(\mathrm{x}+\frac{3}{2}\right)=\mathrm{t} \ldots \ldots$ (i)

$\Rightarrow \mathrm{d} \mathrm{x}=\mathrm{dt}$

SO,

$I=\frac{1}{4} \int \frac{1}{t^{2}-(1)^{2}} d t$

$I=\frac{1}{4} \times \frac{1}{2 \times 1} \log \left|\frac{t-1}{t+1}\right|+c$

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

$I=\frac{1}{8} \log \left|\frac{x-\frac{3}{2}-1}{x+\frac{2}{2}+1}\right|+c$ [using (i)]

$I=\frac{1}{8} \log \left|\frac{2 x-1}{2 x+5}\right|+c$