Integral of exponential function

**emttim84** · January 22nd 2008, 03:03 PM

Ok, so maybe I'm just missing something simple here, but I'm trying to take the integral of x^2 e^x^3 so I'm doing it with the tabular method, so I'm trying to integrate e^x^3dx for dv and for some reason I just can't get it to match up. I'm using a u sub for $x^3$ and then du is $3xdx$ but the original integral, e^x^3dx has no x next to e, so I don't see how I can modify du to match up. Even if you divide by 3 to factor that out, you still have $xdx$ instead of $dx$ .

**Krizalid** · January 22nd 2008, 03:09 PM

Originally Posted by emttim84

Ok, so maybe I'm just missing something simple here, but I'm trying to take the integral of x^2 e^x^3

This is just Chain Rule. (We can even tackle many integrals without applyin' substitution.)

$(x^3)'=3x^2.$ Look at your integrand, you have an $x^2.$

Now

$\int {x^2 e^{x^3 } \,dx} = \frac{1}<br /> {3}\int {\left( {x^3 } \right)'e^{x^3 } \,dx} = \frac{1}<br /> {3}\int {\left( {e^{x^3 } } \right)'\,dx} = \frac{1}<br /> {3}e^{x^3 } + k.$

Differentiate to check.

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