Mathematics

Evaluate $\int \dfrac{1}{{\sqrt {{x^2} + 2x + 2} }}$

SOLUTION
$\int\dfrac{1}{\sqrt{x^2+2x+2}}dx$

$\Rightarrow \int{\dfrac{1}{\sqrt{x^2+2x+1+1}}}dx$

$\Rightarrow \int{\dfrac{1}{\sqrt{(x+1)^2+1}}}dx$

Take $x+1=\tan \theta$          .....$(1)$
$dx=\sec ^2\theta d\theta$

Substitute these values in the above integral, we get,

$\Rightarrow \int{\dfrac{\sec ^2\theta}{\sqrt{\tan ^2\theta+1}}}d\theta$

$\Rightarrow\int\dfrac{\sec ^2\theta}{\sqrt{\sec ^2\theta}}d\theta$              (Since $\sec ^2\theta-\tan ^2\theta=1, \Rightarrow \sec ^2\theta=1+\tan ^2\theta)$

$\Rightarrow \int\dfrac{\sec ^2\theta}{\sec \theta}d\theta$

$\Rightarrow \int{\sec \theta}d\theta$

We know that
$\int{\sec \theta}d\theta=\log|\sec \theta+\tan \theta|+C$            ..........$(2)$

But from $(1)$,  $\tan \theta=x+1$
So, $\sec \theta=\sqrt{1+\tan ^2\theta}$

$\Rightarrow \sec \theta=\sqrt{1+(x+1)^2}$

$\Rightarrow \sec \theta=\sqrt{x^2+2x+2}$

Substituting these values of $\tan \theta$ and $\\sec \theta$ in $(2)$ we get,

$\Rightarrow \int{\sec \theta}=\log|\sqrt{x^2+2x+2}+x+1|+C$

Therefore,

$\int{\dfrac{1}{\sqrt{x^2+2x+2}}}=\log|\sqrt{x^2+2x+2}+x+1|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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