Evaluate these integrals! please!?

Question

|xe^3xdx, |2xcos(3x)dx, |7xln(3x)dx

Answer 1

Noted, I just finished evaluating them/

Answer 2

you need to use integration by parts

so for the first let u=x and dv=e^3xdx, this gives du=dx and v=(e^3x)/3, so you get:

x*(1/3)*e^3x - |(e^3x)/3dx

which is:

1/3xe^3x-1/9e^3x

Answer 3

Use integration by parts for the second and third.

For the second, take u = 2x and dv = cos(3x)dx.

For the third, take u = ln (3x) and dv = 7x dx.

Answer 4

x e^3x dx
integration by parts
Let u = x
du = dx
dv = e^3x dx
v = 1/3(e^3x)
[x e^3x dx] = [u dv] =
uv - [vdu = uv - 1/3[e^3x dx =
uv - 1/9[e^3x] + c
1/3(x)e^3x) - (1/9)(e^3x) + c
2)
[2x cos(3x) dx = 2[x cos(3x)dx
let u = x, du =dx
dv = cos(3x) dx
v = 1/3(sin(3x))
2[x cos(3x) dx = 2[u dv = 2(uv) - 2[v du =
2(uv) - 2(1/3)[sin(3x) dx
2(uv) - 2/3(1/3)(-cos3x) + c
(2/3)(x sin(3x)) + (2/9)(cos 3x) + c

3)
[7x ln(3x) dx = 7[x ln(3x) dx
u = ln(3x) ; du = 1/( x)dx
dv = 7x dx; v = 7x^2/2

[ ln(3x) . 7x dx = [u dv = (uv) - [vdu
(uv) - [7/2(x^2(1/x = (uv) - 7/2[x =
7/2(ln(3x)(x^2)) - 7/2x^2/2 + c =
7/2(ln(3x)(x^2) - 7/4 (x^2) + c