Two integration probs: int e^x(1-e^x)(1+e^x)^10 dx and....

kagan

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Mar 13, 2008
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1
integrate ex(1ex)\displaystyle \int e^{x}(1-e^{x})%20(1+e^{x})^{10}dx

and

integrate (27e9x+e12x)1/3dx\displaystyle \int (27e^{9x} + e^{12x})^{1/3}dx

please help.
 
For the first one, you could let u=1+ex,   du=ex,   u1=ex\displaystyle u=1+e^{x}, \;\ du=e^{x}, \;\ u-1=e^{x}

Make the subs and get:

(2u)u10du\displaystyle \int{(2-u)u^{10}}du

Now, integrate.
 
Hello, kagan!

2)    (27e9x+e12x)13dx\displaystyle 2)\;\;\int \left(27e^{9x} + e^{12x}\right)^{\frac{1}{3}}\,dx

We have:   [e9x(27+e3x)]13  =  (e9x)13(27+e3x)13\displaystyle \text{We have: }\;\left[e^{9x}\left(27 + e^{3x}\right)\right]^{\frac{1}{3}} \;=\;\left(e^{9x}\right)^{\frac{1}{3}}\left(27+e^{3x}\right)^{\frac{1}{3}}

The integral becomes:   e3x(27+e3x)13dx\displaystyle \text{The integral becomes: }\;\int e^{3x}\left(27 + e^{3x}\right)^{\frac{1}{3}}\,dx

\(\displaystyle \text{Let }\,u \;=\;27 + e^{3x} \quad\hdots\quad Got\:it?\)


 
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