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Okay, so I'm working on definite integrals and I calculated the indefinte integral of $\frac{x}{\sqrt{x+1}}$ to be $\frac{2}{3}u^{3/2}-2\sqrt{u}$ where $u=x+1$. The definite integral is on the interval $[0,5]$ so I used my $u$-subsitution equation on the interval $[1,6]$. I keep getting $7.9981$ but I know that's incorrect. What am I doing wrong?

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If $x\in[0,5]$ then $u\in[1,6]$. I'm not sure whether that's a typo or actually your mistake. – Matthew Pressland Dec 11 '12 at 17:47
...$0$ transforms to $1$. Alternatively transform back into $x$ and use the original $[0,5]$. – Jp McCarthy Dec 11 '12 at 17:48
Thats a typo, sorry. It is supposed to be [1,6] – Lizi Dec 11 '12 at 17:48
What are you expecting to get? 6.2223, perhaps? – Rick Decker Dec 11 '12 at 17:53
I figured it out. It was just a calculation error. – Lizi Dec 11 '12 at 17:54
up vote 0 down vote accepted

Using the substitution $u=x+1$, you get $$\int\frac{xdx}{\sqrt{x+1}}=\int\frac{u-1}{\sqrt{u}}du=\int(u^\frac{1}{2}-u^{-\frac{1}{2}})du=\frac{2}{3}u^\frac{3}{2}-2\sqrt{u}$$as you did. Then, to calculate the definite integral, you plug in the following: $$\int_1^5\frac{u-1}{\sqrt{u}}du=\left[\frac{2}{3}u^\frac{3}{2}-2\sqrt{u}\right]_1^6=\frac{4}{3}+2\sqrt{6}=6.23231\cdots$$not $7.9981$.

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Same answer, but with the useful trick of "adding $0$":

\begin{align} \int \frac{x}{\sqrt{x+1}}dx = \int \frac{x+1 -1}{\sqrt{x+1}}&= \int \left(\frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}} \right)dx \\ &= \int\left( (x+1)^{\frac{1}{2}} - (x+1)^{- \frac{1}{2}}\right)dx \\ &= \frac{2}{3} (x+1)^{\frac{3}{2}} - 2 (x+1)^{\frac{1}{2}}. \end{align}

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