Question

Evaluate: $\int \frac{1}{4x^{2}-1}\ dx$

Answer 1

May be its useful for u

Answer 2

Answer:

$\int\:\frac{1}{(4x^{2}-1)}dx=\frac{log\:|2x-1|}{4}-\frac{log\:|2x+1|}{4}+C$

Step-by-step explanation:

This integration can be done by direct formula

i.e.

$\int\:\frac{1}{(2x)^{2}-1^{2}}dx=\frac{1}{(2x+1)(2x-1)}$

now if denominator can be split into than we can integrate

$=\frac{1}{2(2x-1)}-\frac{1}{2(2x+1)} \\ \\ \\=\int\:\frac{1}{2(2x-1)}dx-\int\:\frac{1}{2(2x+1)}dx\\\\\\= \frac{1}{2}\int\:\frac{1}{(2x-1)}dx-\frac{1}{2}\int\:\frac{1}{(2x+1)}dx\\\\\\ = \frac{1}{2}\int\:\frac{1}{(2x-1)}dx=\\\\let\:\:\:2x-1=t\\\\2dx=dt\\\\dx=\frac{dt}{2}\\\\\\=\frac{1}{4}\int\:\frac{1}{t}dt$

$=\frac{log\:t}{4}+C$

by the same way we can calculate

$\frac{1}{2}\int\:\frac{1}{(2x+1)}dx=\frac{log\:t}{4}+C\\\\\\so\\\\\\\int\:\frac{1}{4x^{2}-1}dx=\frac{log\:|2x-1|}{4}-\frac{log\:|2x+1|}{4}+C$