The correct option is A I=x/4a^2(x^2+a^2)^2+3/4a^2{x/2a^2(x^2+a^2)+1/2a^3tan^ 1(x/a)}+C Let I=∫dx/(x^2+a^2)^3 #160; #160; #160; #160; ...

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Property 2

Evaluate ∫d...

Question

Evaluate $\int \frac{d x}{{(x^{2} + a^{2})}^{3}}$ .

I = \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} {\frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{1}{2 a^{3}} {tan}^{- 1} (\frac{x}{a})} + C

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I = \frac{x}{2 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} {\frac{x}{a^{2} (x^{2} + a^{2})} + \frac{1}{a^{3}} {tan}^{- 1} (\frac{x}{a})} + C

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I = \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{1}{4 a^{2}} {\frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{1}{2 a^{3}} {sec}^{- 1} (\frac{x}{a})} + C

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I = \frac{x}{a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{2 a^{2}} {\frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{1}{2 a^{3}} {tan}^{- 1} (\frac{x}{a})} + C

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Solution

The correct option is A $I = \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} {\frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{1}{2 a^{3}} {tan}^{- 1} (\frac{x}{a})} + C$
Let $I = \int \frac{d x}{{(x^{2} + a^{2})}^{3}}$ ....(i)
and $I_{1} = \int \frac{1}{{(x^{2} + a^{2})}^{2}} d x$ ......(ii)
$I_{1} = \int \frac{1}{{(x^{2} + a^{2})}^{2}} .1 d x$
$= \frac{1}{{(x^{2} + a^{2})}^{2}} . x - \int \frac{- 2 (2 x)}{{(x^{2} + a^{2})}^{3}} x d x$
$= \frac{x}{{(x^{2} + a^{2})}^{2}} + 4 \int \frac{x^{2} + a^{2} - a^{2}}{{(x^{2} + a^{2})}^{3}} d x$
$= \frac{x}{{(x^{2} + a^{2})}^{2}} + 4 \int \frac{1}{{(x^{2} + a^{2})}^{2}} d x - 4 a^{2} \int \frac{d x}{{(x^{2} + a^{2})}^{3}}$
$⟹ I_{1} = \frac{x}{{(x^{2} + a^{2})}^{2}} + 4 I_{1} - 4 a^{2} . I$ [using equations (i) and (ii)]
$⟹ 4 a^{2} . I = \frac{x}{{(x^{2} + a^{2})}^{2}} + 3 I_{1}$
$⟹ I = \frac{x}{{4 a^{2} (x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} I_{1}$ (iii)
$⟹ I = \frac{x}{4 a^{2} {(x^{2} + a^{2})}^{2}} + \frac{3}{4 a^{2}} {\frac{x}{2 a^{2} (x^{2} + a^{2})} + \frac{1}{2 a^{3}} {tan}^{- 1} (\frac{x}{a})} + C$

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