$\displaystyle \int {\ln (\sqrt x + \sqrt {1 + x} )dx} $ I am thinking possibly a u sub but I have no idea what I could sub in.
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Originally Posted by genlovesmusic09 $\displaystyle \int {\ln (\sqrt x + \sqrt {1 + x} )dx} $ I am thinking possibly a u sub but I have no idea what I could sub in. Try by parts with $\displaystyle u=\ln(\sqrt{x}+\sqrt{x+1})$ $\displaystyle dv=1dx$
would the derivative of u=ln(sqrtx + sqrt1+x) be: $\displaystyle \frac{{\frac{1} {{2\sqrt x }} + 1}} {{x + \sqrt x + 1}} $ and if so how would i integrate that (timeing it by v=x)?
Originally Posted by VonNemo19 Try by parts with $\displaystyle u=\ln(\sqrt{x}+\sqrt{x+1})$ $\displaystyle dv=1dx$ what would du be?
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