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IfI=int(dx)/((2a x+x^2)^(3/2)),then I is...

`IfI=int(dx)/((2a x+x^2)^(3/2))`,then `I` is equal to`` (a)`-(x+a)/(sqrt(2a x+x^2))+c` (b) `-1/a(x+a)/(sqrt(2a x+x^2))+c` (c)`-1/(a^2)(x+a)/(sqrt(2a x+x^2))+c` (d) `-1/(a^3)(x+a)/(sqrt(2a x+x^3))+c`

Text Solution

Verified by Experts

The correct Answer is:
`-(x+a)/(a^(2)sqrt(2ax+x^(2)))+C`
`2ax+x^(2)=(x+a)^(2)-a^(2).`
Now, put `x+a=a sec theta`
` :. dx= a sec theta tan theta d theta`
` :. int (dx)/((2ax+x^(2))^(3//2))=int(a sec theta tan theta)/(a^(3) tan^(3) theta)d theta`
`=(1)/(a^(2))int(cos theta)/(sin^(2) theta) d theta`
`= -(1)/(a^(2)sin theta)+C`
`= -(1)/(a^(2)sqrt(1-(a^(2))/((x+a)^(2))))+C`
`=-(x+a)/(a^(2)sqrt(2ax+x^(2)))+C`
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