∫e^x(3e^x + 1)^1/3 dx = Can anyone solve this problem?

Question

(3/5)(3e^x + 1)^5/3 + C ?

(2/9)(3e^x + 1)^3/2 + C ?

Answer 1

Let u be 3e^x + 1, then du/dx = 3e^x . So your integral amounts to

integral( (1/3 du/dx) u^(1/3)) = 1/3 integral(u^(1/3)du)

= (1/3)*1/(1+(1/3))*u^(1 + 1/3) + C
=(1/3)*(1/(4/3))*u^(4/3) + C
= (1/3) * (3/4) * (3e^x+1)^(4/3) + C
= (1/4)* (3e^x + 1)^(4/3) + C

Check by differentiating:

d/dx (1/4 (3e^x + 1)^(4/3) + C) =
(4/3)* (1/4) * (3e^x + 1)^(1/3) * (d/dx (3e^x + 1)) + d/dx(C) =
(4/12) * (3e^x + 1)^ (1/3) * 3e^x dx + 0 + 0 =
(1/3) * (3e^x + 1)^(1/3) * 3e^x dx =
e^x (3e^x + 1) dx

which is what you started with.

Answer 2

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

Answer 3

(1/3)(e^x + (3/2)e^(2x)) + C

Answer 4

=c9(&^0)"xx B7/34cXZ9 ok HAPY