Integration by parts

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krazydog

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Oct 25, 2011
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I have a problem where i am not sure how the solutions manual got the step. I understand they are taking
integration by parts twice but i dont see how they are producing the 6 in the second integration.

\(\displaystyle \int x^3e^x dx\)

\(\displaystyle u = x^3 , dv = e^x\)

Take integration by parts 3 times..
\(\displaystyle u = x , v = e^x, du =dx\)

Now, lets put it in the form \(\displaystyle \int u dv = uv - \int v du\)

We have
\(\displaystyle \int x^3 e^x dx = x^3e^x -3 \int x^2e^x dx\)


Now from here we need to do integration by parts on the integral again because it doesnt fit any rules?
But when i do

\(\displaystyle u = x^2 , dv = e^x\)
So, then i would have \(\displaystyle du = 2xdx , v = e^x\)

now, this is where i am not sure how they get 6?

\(\displaystyle = x^3e^x - 3x^2e^x + 6 \int xe^x dx \)

Do i need to distribute the -3 to the integral and times it by the 2xdx?
 
Hello, krazydog!

If you write out all the steps, you shouldn't get lost . . .

\(\displaystyle \displaystyle I \;=\;\int x^3e^x\,\!dx\)
Click to expand...

By parts: .\(\displaystyle \begin{Bmatrix}u &=& x^3 && dv &=& e^x\,\!dx \\ du &=& 3x^2\,\!dx && v &=& e^x \end{Bmatrix}\)

Then: .\(\displaystyle \displaystyle I \;=\;x^3e^x - 3\!\!\int\!\! x^2e^x\,\!dx\)

By parts: .\(\displaystyle \begin{Bmatrix}u &=& x^2 && dv &=& e^x\,\!dx \\ du &=& 2x\,dx && v &=& e^x \end{Bmatrix}\)

Then: . \(\displaystyle \displaystyle I \;=\;x^3e^x - 3\left[x^2e^x\,-\,2\!\!\int\!\! xe^x\,dx\right] \)

. . . . . .\(\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6\!\!\int\!\! xe^x\,dx \)

By parts: .\(\displaystyle \begin{Bmatrix}u &=& x && dv &=& e^x\,dx \\ du &=& dx && v &=& e^x \end{Bmatrix}\)

Then: .\(\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6\left[xe^x - \int\!\!e^x\,dx\right] \)

. . . . . \(\displaystyle \displaystyle I \;=\;x^3e^x - 3x^2e^x + 6xe^x - 6\!\!\int\!\! e^x\,dx\)

. . . . . \(\displaystyle I \;=\;x^3e^x - 3x^2e^x + 6xe^x - 6e^x + C\)


. . . . . \(\displaystyle I \;=\;e^x(x^3 - 3x^2 + 6x - 6) + C\)
 
Yes, it comes from the 3 from before being multiplied by the 2 in the last step.

But, here is a shorter way to do this problem.

Since there is a cubic term in the integrand, the solution will have the form

\(\displaystyle (ax^{3}+bx^{2}+cx+d)e^{x}\)


Differentiate using the product rule:

\(\displaystyle (ax^{3}+bx^{2}+cx+d)e^{x}+e^{x}(3ax^{2}+2bx+c)e^{x}=x^{3}e^{x}\)

Equate coefficients:

\(\displaystyle a=1, \;\ 3a+b=0, \;\ 2b+c=0, \;\ c+d=0\)

a is already solved, and this leads to \(\displaystyle a=1, \;\ b=-3, \;\ c=6, \;\ d=-6\)

\(\displaystyle (x^{3}-3x^{2}+6x-6)e^{x}\)
 
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