Mathematics

# Solve $\displaystyle\int\dfrac{x}{{\sqrt {{a^2} + {x^2}} }}dx$

##### SOLUTION

Consider the given integral.

$I=\displaystyle\int{\dfrac{x}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx}$

Let $t={{a}^{2}}+{{x}^{2}}$

$\dfrac{dt}{dx}=0+2x$

$\dfrac{dt}{2}=xdx$

Therefore,

$I=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}$

$I=\dfrac{1}{2}\times 2\sqrt{t}+C$

$I=\sqrt{t}+C$

On putting the value of $t$, we get

$I=\sqrt{{{a}^{2}}+{{x}^{2}}}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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