Im having trouble finding the antiderrivative of r/[sqrt(4+r^2)] the answer is [sqrt(4+r^2)]...but how can that be?
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Originally Posted by johntuan Im having trouble finding the antiderrivative of r/[sqrt(4+r^2)] the answer is [sqrt(4+r^2)]...but how can that be? $\displaystyle \int \frac{r}{\sqrt{4+r^2}}dr$ let $\displaystyle u=4+r^2 \implies du=2rdr \iff \frac{du}{2}=rdr$ $\displaystyle \int \frac{r}{\sqrt{4+r^2}}dr=\frac{1}{2}\int \frac{1}{u^{1/2}}du=... $
but then wouldnt the answer be over 2? u^(1/2)/2 but the answer doesnt have a denominator:s
Not quite $\displaystyle \frac{1}{2}\int \frac{1}{u^{1/2}}du=\frac{1}{2}\int u^{-1/2}du=\frac{\frac{1}{2}}{\frac{1}{2}}u^{1/2}=\sqrt{u}=\sqrt{4+r^2}$
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