Mathematics

The value of $$\displaystyle \int {\dfrac{{d({x^2} + 1)}}{{\sqrt {{x^2} + 2} }}} ,$$ is 


ANSWER

$$2\sqrt {{x^2} + 2} + c$$


SOLUTION
Consider $$I=\displaystyle \int \dfrac{d(x^2+1)}{\sqrt{x^2+2}}$$ we know $$d(x^2+1)=2x\ dx$$ 

$$T=\displaystyle \int \dfrac{2x\ dx}{\sqrt{x^2+2}}$$ put $$x^2 + 2 = t \Rightarrow 2x\ dx = dt $$

$$I=\displaystyle \int \dfrac{dt}{\sqrt t} = 2\sqrt{t}+c \Rightarrow 2\sqrt{x^2+2}+ c.$$
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Single Correct Medium Published on 17th 09, 2020
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