K kagan New member Joined Mar 13, 2008 Messages 1 Mar 13, 2008 #1 integrate \(\displaystyle \int e^{x}(1-e^{x})%20(1+e^{x})^{10}dx\) and integrate \(\displaystyle \int (27e^{9x} + e^{12x})^{1/3}dx\) please help.
integrate \(\displaystyle \int e^{x}(1-e^{x})%20(1+e^{x})^{10}dx\) and integrate \(\displaystyle \int (27e^{9x} + e^{12x})^{1/3}dx\) please help.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Mar 13, 2008 #2 For the first one, you could let \(\displaystyle u=1+e^{x}, \;\ du=e^{x}, \;\ u-1=e^{x}\) Make the subs and get: \(\displaystyle \int{(2-u)u^{10}}du\) Now, integrate.
For the first one, you could let \(\displaystyle u=1+e^{x}, \;\ du=e^{x}, \;\ u-1=e^{x}\) Make the subs and get: \(\displaystyle \int{(2-u)u^{10}}du\) Now, integrate.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Mar 13, 2008 #3 Hello, kagan! \(\displaystyle 2)\;\;\int \left(27e^{9x} + e^{12x}\right)^{\frac{1}{3}}\,dx\) Click to expand... \(\displaystyle \text{We have: }\;\left[e^{9x}\left(27 + e^{3x}\right)\right]^{\frac{1}{3}} \;=\;\left(e^{9x}\right)^{\frac{1}{3}}\left(27+e^{3x}\right)^{\frac{1}{3}}\) \(\displaystyle \text{The integral becomes: }\;\int e^{3x}\left(27 + e^{3x}\right)^{\frac{1}{3}}\,dx\) \(\displaystyle \text{Let }\,u \;=\;27 + e^{3x} \quad\hdots\quad Got\:it?\)
Hello, kagan! \(\displaystyle 2)\;\;\int \left(27e^{9x} + e^{12x}\right)^{\frac{1}{3}}\,dx\) Click to expand... \(\displaystyle \text{We have: }\;\left[e^{9x}\left(27 + e^{3x}\right)\right]^{\frac{1}{3}} \;=\;\left(e^{9x}\right)^{\frac{1}{3}}\left(27+e^{3x}\right)^{\frac{1}{3}}\) \(\displaystyle \text{The integral becomes: }\;\int e^{3x}\left(27 + e^{3x}\right)^{\frac{1}{3}}\,dx\) \(\displaystyle \text{Let }\,u \;=\;27 + e^{3x} \quad\hdots\quad Got\:it?\)
