James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
Hi there, I am stuck on this integration problem and would be very appreciative of some guidance. I have tried a few things now and my methods do not seem to reach a satisfactory conclusion. I think the solution involves some sort of combination of solving by substitution and integration by parts but apparently I cannot work out the right method. Here is the problem and what I have tried:
Find
[math]\int x^3 \ln {(x^2 +1)} \ dx[/math]
let's start with the substitution method,
[math]u = x^2+1[/math]
now you can re-write the problem:
[math]\int\frac{(u-1)}{2}\ln{u} \:du[/math]
[math]\frac{1}{2}\int(u-1)\ln{u} \:du[/math]
which looks like it could be amenable to the method of integration by parts:
[math]u = u-1 \\du = 1[/math]
[math]v = u \ln{u} - u \\ dv = \ln{u}[/math]
[math]I = (u-1)(u\ln{u} - u) - \int u\ln{u} - u[/math]
which gives rise to a successive integration by parts, and the problem becomes circular.
I think this is more complicated than it should be, is there something else I should try?
Find
[math]\int x^3 \ln {(x^2 +1)} \ dx[/math]
let's start with the substitution method,
[math]u = x^2+1[/math]
now you can re-write the problem:
[math]\int\frac{(u-1)}{2}\ln{u} \:du[/math]
[math]\frac{1}{2}\int(u-1)\ln{u} \:du[/math]
which looks like it could be amenable to the method of integration by parts:
[math]u = u-1 \\du = 1[/math]
[math]v = u \ln{u} - u \\ dv = \ln{u}[/math]
[math]I = (u-1)(u\ln{u} - u) - \int u\ln{u} - u[/math]
which gives rise to a successive integration by parts, and the problem becomes circular.
I think this is more complicated than it should be, is there something else I should try?