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
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