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
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