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1/((sqrtx)(sqrtx+1)) dx integral
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is it (sqrtx +1) or sqrt(x+1)?
\[1/((\sqrt{x})(\sqrt{x}+1))\]
the answer from the book is\[2\ln \left| \sqrt{x}+1 \right|+c\]
this is equal to dx/(x + sqrtx) . let u = sqrtx so du = 1/2 (x^(-1/2)) * dx this is equal to [1/(u^2 + u)] * 2u du = 2u/(u^2 + u) lmk if you would like the next step
ok i substituted sqrtx back in but its not the book answer
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ooh the book say use (sqrtx+1)=u
2u/(u^2 + u) = 2/(u +1) --> simpler to integrate
both substitutions work. in fact even a whacky substitution like u = sqrtx + pi could work if done right. integral is 2 ln |u + 1| + C . u = sqrtx. ^_^
thanks you so much for helping me
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