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`1f I=int(dx)/((2a x+x^2)^(3/2))` |
Answer» Correct Answer - `-(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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