Integration in calc two

Question

How do you integrate ∫4x^3e^(x^2 - 3) dx

Answer 1

By parts.

First you choose a u (which gets differentiated) and a dv/dx (which gets integrated).

If we choose x³ to get integrated, we'd repeatedly increment the exponent of x without getting anywhere. If we choose e^x²-3 to be integrated, well, that function doesn't have an elementary integral.

But, there are other ways to factor that. For example, even though e^x²-3 doesn't have a nice clean integral, its derivative, 2xe^x²-3, does. And our expression most certainly does factor out xe^x²-3 (the 2 is just a constant out front).

So, we choose dv/dx = 2xe^x²-3, and u = 2x².

∫ u * dv/dx dx = u*v - ∫ du/dx * v dx

∫ 2x² * 2xe^x²-3 dx = 2x² * e^x²-3 - ∫ 4x * e^x²-3 dx = 2x²e^x²-3 - 2 * e^x²-3 + C

So ∫ 4x³e^x²-3 dx = 2(x²-1)e^x²-3 + C

Just to check, let's differentiate it:

d/dx 2(x²-1)e^x²-3 + C = 2(2x)(e^x²-3) + 2(x²-1)(2xe^x²-3) + 0 = (4x+4x³-4x)e^x²-3 = 4x³e^x²-3

Yup, it works out.

Answer 2

4∫x³e^{(x^2 - 3)}dx = 4∫x²xe^{(x^2-3)}dx

Let u = x². Then du = 2xdx. Let dv = xe^{(x^2 - 3)}dx. Then v = ∫dv = (1/2)e^{(x^2 - 3)}.

So, 4∫x³e^{(x^2 - 3)}dx = 4[uv - ∫vdu] = 4[ (1/2)x²e^{(x^2-3)} - ∫xe^{(x^2-3)}dx] = 2x²e^{(x^2-3)} - 2e^{(x^2-3)} + C

= 2(x² - 1)e^{(x^2-3)} + C