# Integrate $x(2x+5)^{10}$ by substitution

To find $\int x(2x+5)^{10} dx$ I started to set $u=2x+5$. Then $du=2 \space dx$ and $x=\frac{u-5}{2}$.

So,

$$\int x(2x+5)^{10}dx=$$ $$=\frac{1}{2}\int x \space u^{10}2 \space dx=$$ $$=\frac{1}{2}\int \frac{u-5}{2} \space u^{10}2 \space dx=$$ $$=\frac{1}{4}\int (u-5)u^{10} du=$$ $$\frac{1}{4}\int u^{11}-5u^{10} du=$$ $$=\frac{1}{4}\int u^{11}du-\frac{5}{4}\int u^{10}du=$$

Then one gets,

$$=\frac{1}{4}\cdot\frac{u^{12}}{12}-\frac{5}{4}\cdot\frac{u^{11}}{11}=$$ $$=\frac{u^{12}}{48}-\frac{5u^{11}}{44}$$ $$\frac{(2x+5)^{12}}{48}-\frac{5(2x+5)^{11}}{44}$$

I know that more algebra is need to get the final result. But, is my thought correct? Thanks.

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I recommend never writing $u$ and $x$ in the same integral when doing substitution. While there may be no rule against it, when I see students do it, they always get the wrong answer. –  Stefan Smith Apr 19 '13 at 21:50

You can check your result by differentiation: $$\frac{d}{dx} [ \frac{ 11 \cdot (2x+5)^{12} - 12 \cdot 5 \cdot (2x+5)^{11} }{4 \cdot 11 \cdot 12} ]$$

$$= \frac{ [11 \cdot 12 \cdot (2x+5)^{11} \cdot 2] -[ 12 \cdot 5 \cdot 11 \cdot (2x+5)^{10} \cdot 2] }{4 \cdot 11 \cdot 12}$$

$$= (2x+5)^{10} \cdot \frac{ [11 \cdot 12 \cdot 2 \cdot (2x+5)] -[ 12 \cdot 5 \cdot 11 \cdot 2 ] }{4 \cdot 11 \cdot 12}$$

$$= (2x+5)^{10} \cdot \frac{ (11 \cdot 12 \cdot 2 \cdot 2x) + (5 \cdot 11 \cdot 12 \cdot 2 ) - (12 \cdot 5 \cdot 11 \cdot 2) }{4 \cdot 11 \cdot 12}$$

$$= (2x+5)^{10} \cdot \frac{ 11 \cdot 12 \cdot 2 \cdot 2x }{4 \cdot 11 \cdot 12} = (2x+5)^{10} \cdot x$$

While Stano is formally correct (and some graders might actually penalize you for writing things that way in an exam solution), I frequently see people put the two different variables under the same integral sign in calculation work. It can be understood as a "shorthand" for the integration still being with respect to $x$, but "$u$" stands for $u(x)$.

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Yes, your thought is correct, but there are some mistakes $\frac{1}{2}\int \frac{u-5}{2} \space u^{10}2 \space d\color{red}x$ (first and second line)

The notation: $\frac{1}{2}\int x \space u^{10}2 \space dx$ is also not correct because you mix variables $u$ and $x$ in one integral

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Integrate x(2x+5)"10dx...........let u=2x+5. Therefore, dx=du/2 and x=(u-5)/2........int:x(2x+5)"10dx=int:{(u-5)/2}(u)"2du/2........=int:(u-5)(u)"2/4du=int:(u"11-5u"10)du/4.........=u"12/16-5u"11/12+c where c is the constant of integration. =(2x+5)"12/16-5(2x+5)"11/12+c...............int:{(2x+5)"11/12}{(2x+5)/4-5/11}+c............int:{(2x+5)"11/12}{x/2-5/44}+c.............=(2x+5)"11(22x-5)/528+C. It could be solved also by expanding the expression. or using integration by parts treating x as first function and (2x+5)"10 as second function. Hope this help......... By O'john

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Please use Mathjax to format your answer. See here for details of how to use Mathjax. –  Peter Phipps Jan 16 '14 at 11:36