Integration by parts expression help

AStaunton
Messages
100
Reaction score
1
the expression to integrate is:

\int x^{3}e^{x^{2}}dx

and in the spirit of "LIATE" I set my u and dv as the following:

dv=e^{x^{2}}dx

u=x^{3}

however, doing this that I integrate dv=e^{x^{2}}dx in order to get v...and unless I'm missing something, this does not seem like an easy integral! a u substition won't work as i'd need an x^1 term multiplying by the e^x^2...

and going the other way and setting dv=x^3dx and u=e^x^2 and plugging into int,by parts formula gets:

\frac{x^{4}}{4}e^{x^{2}}-\int\frac{x^{4}}{4}2xe^{x^{2}}dx

and I don't think further integration by parts will help with this new integral..

any advice appreciated.
 
Physics news on Phys.org
correction of typo:



however, doing this REQUIRES that I integrate dv=e^{x^{2}}dx in order to get v...
 
change variable x^2->u
 
Perform a u-substitution with u = x3, and then integrate by parts.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Back
Top