Mathematics

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


SOLUTION

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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