Integral problem involving natural log?

Question

integral of ln (x^2+1) dx. with lower bound 0, upper bound e^x. Thanks in Advance.

Answer 1

Given function is

Let dv = dx

v = x

Let u = ln (x²+1) and

du = [1/(x²+1)]2xdx

du = [1/(v²+1)]2vdv

du = [2v/(v²+1)]dv

d(uv) = udv + vdu

Apply integration both sides.

∫ d(uv) = ∫ ( udv + vdu )

uv = ∫ udv + ∫ vdu

∫ udv = uv - ∫ vdu

Now Substitute : u = ln (x²+1) , v = x , du = [2v/(v²+1)]dv  and dv = dx

= x ln (x²+1) - 2v + 2tan^-1(v) + c

Substitute v = x

= x ln (x²+1) - 2x + 2tan^-1(x)

Now substitute the limits

The lower bound = 0

The upper bound  = e^x

= [ (e^x) ln ((e^x)²+1) - 2(e^x) + 2tan^-1(e^x) ]  - [ 0 ln (0²+1) - 2*0 + 2tan^-1(0) ]

= [ (e^x) ln (e^2x+1) - 2e^x + 2tan^-1(e^x) ]  - [ 0 - 0 + 0 ]

= (e^x)ln (e^2x+1) - 2e^x + 2tan^-1(e^x)

Solution :

⌠ ln (x²+1) dx = (e^x)ln (e^2x+1) - 2e^x + 2tan^-1(e^x)