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$$\int^{1}_{0} x^2(x+1)^5 dx$$

I would appreciate if you could go through the process step by step. I have been working on this problem for quite awhile and cannot seem to figure it out.

after finding the $u, du, dv, v$ my equation is

$\frac{x^2(x+1)^6}{6}-$ $\int^{1}_{0} \frac{ (x+1)^6}{6 (2x)}dx$

Math137
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user149420
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    http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference –  May 11 '14 at 22:07
  • You're on the right track. You just need to do integration by parts one more time with that second integral, then you'll be left with a third integral you can solve without integration by parts. (To make your work a bit more organized, you should try the tabular method) – user137794 May 11 '14 at 22:07
  • @user149420 is that what you wanted to write? – Math137 May 11 '14 at 22:14

2 Answers2

1

Your first part is correct, your second is not. It should be

$$\int vdu=\int\frac{(x+1)^6}62xdx$$

Another round of integration by parts will eliminate that last $x$. However, it's probably a lot easier to solve the problem without integration by parts. Just a simple substitution

$$u=x+1,du=dx$$ $$\int_1^2(u-1)^2u^5du=\int_1^2u^7-2u^6+u^5du$$

Mike
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0

Do integration by parts one more time.

$$\int x^2(x+1)^5\ dx \\ = x^2\cdot\frac16(x+1)^6-\int2x\cdot\frac16(x+1)^6\ dx \\ = x^2\cdot\frac16(x+1)^6-2x\cdot\frac1{42}(x+1)^7+\int\frac1{21}(x+1)^7\ dx $$

user137794
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