Mathematics

# $\int { \dfrac { 1 }{ \left( 4x-1 \right) \sqrt { { \left( 4x-1 \right) }^{ 2 }-1 } } } dx =$

## Consider the following question.

$I=\int\limits_{{}}^{{}}{\dfrac{1}{\left( 4x-1 \right)\sqrt{{{\left( 4x-1 \right)}^{2}}-1}}}dx\,\,\,\,\,\,...........\left( 1 \right)$

Let, and differentiate w.r.t to dx we get.

$t=\left( 4x-1 \right)$

$\dfrac{dt}{dx}=4$

$=\dfrac{1}{4}\int\limits_{{}}^{{}}{\dfrac{1}{t\sqrt{{{t}^{2}}-1}}}dt$

$=\dfrac{1}{4}{{\sec }^{-1}}\left( t \right)$

$=\dfrac{1}{4}{{\sec }^{-1}}\left( 4x-1 \right)+C$

Hence, this is the required answer.

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Subjective Medium Published on 17th 09, 2020
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