Mathematics

Evaluate
$$\displaystyle \int \dfrac {x+2}{\sqrt {x^{2}+4x+1}}.dx$$


SOLUTION
We have,
$$I=\displaystyle \int \dfrac {x+2}{\sqrt {x^{2}+4x+1}}dx$$

Let
$$t=x^2+4x+1$$
$$\dfrac{dt}{dx}=2x+4$$
$$\dfrac{dt}{2}=(x+2)dx$$

Therefore,
$$I=\dfrac{1}{2}\displaystyle \int \dfrac {1}{\sqrt {t}}dt$$
$$I=\dfrac{1}{2}(2\sqrt t)+C$$
$$I=\sqrt t+C$$

On putting the value of $$t$$, we get
$$I=\sqrt {x^2+4x+1}+C$$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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