Mathematics

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

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

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

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