how do u inegrate inegrate: e^x(2x)^2 dx?

Question

its calculus 2
how do i integrate that proble

Answer 1

Use integration by parts:
integral(udv) = uv - integral(vdu) [1]

You have the function e^x multiplied by (2x)^2.

Pick u = (2x)^2 (easy derivative)
Pick dv = e^x (easy integral)

Then:

du = 8x
v = e^x.

Substituting it all in the Integration by parts formula[1]:

integral(e^x(2x)^2) = e^x(2x)^2 - integral(e^x(8x)) [2]

Now need to solve integral(e^x(8x))

Again use integration by parts where:

u = 8x (easy derivative)
dv = e^x (easy integral)

Then:

du = 8
v = e^x

Therefore:

integral(e^x(8x)) = e^x(8x) - integral(e^x(8))
= e^x(8x) - e^x(8)

Substituting it in to formula[2]:

integral(e^x(2x)^2) = e^x(2x)^2 - e^x(8x) + e^x(8) + C

Answer 2

You have to do integration by parts... twice.
I don't really like doing peoples hw, but this is how you do it.

u= 2x^2 dv = e^x dx
du= 4x dx v=e^x

Substituting into uv - (integral) vdu you get:

e^x 2x^2- (integral) e^x 4x dx

Integrate by parts again. You use the same method with

u= 4x dv = e^x
du = 4 dx v= e^x

Substitute again you get

e^x 2x^2- (4x e^x - (integral) 4 e^x dx)

You can probably take it from here. Enjoy.


I just saw the other persons answer. I'm not sure if you mean
(2x)^2 or 2x^2. If it's the first one, the other person has it right, if it's the latter, mine is right.

Answer 3

use the i think its name is integration by parts. the fromula is,

integaral udv = uv - integral vdu

let u = (2x)^2 = 4x^2 and dv = e^xdx, so du = 8xdx and v = e^x

inetagral (4x^2)(e^xdx) = (4x^2)(e^x) - integarl (e^x)(8xdx)

again use this method of integration to find integarl (e^x)(8xdx). yhis time let u =8x and dv = e^xdx, so du=8 and v=e^x.

and solve it.

the trick is u must use this method twice not once

Answer 4

Use the formula I=integration symbol
Iu.dv=u.v-I v.du