Mathematics

The value of $$\displaystyle \int \dfrac {x^2+1}{\sqrt {x^2 +2} }dx$$ is equal to:


ANSWER

$$\dfrac{x\sqrt {({x^2} + 2)}}{2} + C$$


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

$$=\displaystyle \int \dfrac {x^2+2-1}{\sqrt {x^2+2} }dx$$

$$=\displaystyle\int{\sqrt {x^2+2} }dx$$$$-\displaystyle \int \dfrac {1}{\sqrt {x^2+2} }dx$$

$$=\dfrac{x\sqrt{x^2+2}}{2}+\dfrac{2}{2}\log\left |x+\sqrt{x^2+2}\right|-\log\left|x+\sqrt{x^2+2}\right|+c$$ [ Where $$c$$ is integrating constant]

$$=\dfrac{x\sqrt{x^2+2}}{2}+c$$.
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Single Correct Medium Published on 17th 09, 2020
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