How do I solve this integral..?

I = ∫ [(4xe^(6x^2)] dx

u = 6x^2

du = 12x dx

(1/3)du = 4x dx

I = (1/3)∫ e^u du

I = (1/3)e(6x^2) + C<-------answer

Substitute 6x^2 as t.

Your integral looks like

∫ 4xe^t dx

On differentiating t=6x^2 you will get dt = 12x dx

dx = dt / 12x

Substitute this and your integral looks like

∫ 1/3 e^t dt

The 1/3 is a constant take it out of the integral sign

1/3 ∫ e^t dt

You know that ∫ e^x dx = e^x + c (where c is an arbitrary constant)

1/3 ∫ e^t dt = 1/3 e^t + c

Re-substituting t

1/3 e^(6x^2) + c

Use substitution

Let u=6x^2 so du=12xdx so dx=du/12x

Now replace u and du.

∫ [(4xe^(6x^2)] dx= ∫ 4xe^u(du/12x)= ∫1/3e^udu=1/3∫e^udu=1/3e^u

Now substitute for u

so 1/3e^u= 1/3e^(6x^2) +C

so 1/3e^(6x^2) +C is the answer

pull the 4 out in front let u=6x^2 du= 12x integrate

1/3 integral e^u du = 1/3 (e^u) = 1/3 e^(6x^2) + c

I'm going to say u substitution,

u=6x^2

du=12xdx

then you get

1/3e^6x^2+c

after integrating

and wolfram alpha confirms

solve with integration by parts

set u=4x and dv=e^(6x^2)

du=4dx and v=[e^(6x^2)}/(6x^2)

and then take

uv-∫vdu

(something like that)

obvious substitution.

Too hard!