Help with Integrating: xln(1+x)dx using substitution.

I am having a difficult time working this Integral. The instructions are to first make a substitution and then integrate by parts. I understand both, but I feel this one has me working in circles. Any insight would be appreciated.

The indefinite integral (most general) of xln(1+x)dx

Re: Help with Integrating: xln(1+x)dx using substitution.

Make the substitution u = (x+1). This will leave you with:

(u-1)ln(u). Deal separately with u*ln(u) and -ln(u). Each of them can be integrated using integration by parts. In the first case, integrate u and differentiate ln(u); the product of this will be a linear function easy to integrate. The second one is similar, write it as -1 * ln(u). I'll leave the details to you, tell me if this was too vague.

Re: Help with Integrating: xln(1+x)dx using substitution.

How about:

Re: Help with Integrating: xln(1+x)dx using substitution.

Let me show you what I've been trying.

I don't see how to make an integral sign. So I will use a normal 'S'

S xln(1+x)dx Let t = 1 + x then x = t - 1

so dt = (1/2)x^{2}dx and dx = (2/x^{2})dt

so in terms of t:

S (t - 1)ln(t)(2/(t-1))^{2}dt = S [2(t - 1) / (t - 1)^{2}]ln(t)dt = 2S [1 / (t - 1)ln(t)dt

everything I get after this seems wrong. No matter how I manipulate it. Did I substitute incorrectly?

Re: Help with Integrating: xln(1+x)dx using substitution.

Quote:

Originally Posted by

**iAmKrizzle** Let me show you what I've been trying.

I don't see how to make an integral sign. So I will use a normal 'S'

S xln(1+x)dx Let t = 1 + x then x = t - 1

so dt = (1/2)x^{2}dx and dx = (2/x^{2})dt

so in terms of t:

S (t - 1)ln(t)(2/(t-1))^{2}dt = S [2(t - 1) / (t - 1)^{2}]ln(t)dt = 2S [1 / (t - 1)ln(t)dt

everything I get after this seems wrong. No matter how I manipulate it. Did I substitute incorrectly?

Last time I checked, if t = 1 + x, then dt/dx = 1 and so dt = dx...

Re: Help with Integrating: xln(1+x)dx using substitution.

Hello, iAmKrizzle!

I don't know why they suggested a substitution first.