Why Does My Integration by Parts Result Differ?

[greg_rack]

Homework Statement

$$\int \frac{lnx}{(x+1)^2} dx$$

Relevant Equations

none

Hi guys,

I've attempted to integrate this function by parts, which seemed to be the most appropriate method... but apparently, I'm getting something wrong since the result doesn't match the right one.
Everything looks good to me, but there must be something silly missing :)
My attempt:

 

[PeroK]

You seem to have gone wrong at the end. There is nothing you can do with $\frac{\ln x}{x+1}$.

 

[greg_rack]

You seem to have gone wrong at the end. There is nothing you can do with $\frac{\ln x}{x+1}$.

You mean I have gone wrong by collecting $lnx$ at the very end?

 

[PeroK]

You mean I have gone wrong by collecting $lnx$ at the very end?

Yes, the last three steps seem unnecessary and you've lost the term $\ln|x +1|$ somehow.

 

[greg_rack]

Yes, the last three steps seem unnecessary and you've lost the term $\ln|x +1|$ somehow.

Got it, that's actually taking me to nowhere... but still, I can't understand how to reach the solution
$$\frac{xlnx}{x+1}-ln(x+1)+c$$ provided by my textbook.
Is it just a simplification(that apparently I can't see) far from what I've got, or have I got something wrong in integrating?

 

[PeroK]

Got it, that's actually taking me to nowhere... but still, I can't understand how to reach the solution
$$\frac{xlnx}{x+1}-ln(x+1)+c$$ provided by my textbook.
Is it just a simplification(that apparently I can't see) far from what I've got, or have I got something wrong in integrating?

That combines the two terms you have in $\ln x$.

PS Although it looks like they got the sign wrong.

 

[greg_rack]

That combines the two terms you have in $\ln x$.

Uhm, but how? I could add the two terms in $\ln x$, and still end up with a useless simplification

 

[PeroK]

Uhm, but how? I could add the two terms in $\ln x$, and still end up with a useless simplification

The answer's wrong. They got the sign wrong on the parts, I suspect.

 

[PeroK]

PS You can always check an indefinite integral by differentiating, of course. That's something everyone should know!

 

[greg_rack]

PS You can always check an indefinite integral by differentiating, of course. That's something everyone should know!

And indeed, differentiating my result, the integration appears to be incorrect, while the book's one is right... where did I get it wrong?

 

[PeroK]

And indeed, differentiating my result, the integration appears to be incorrect, while the book's one is right... where did I get it wrong?

You got the sign wrong, then! Right at the start.

 

[Fred Wright]

Instead of integration by parts I suggest making the substitution $u=\frac{1}{x+1}$, $du=\frac{-dx}{(x+1)^2}$.

 

[greg_rack]

You got the sign wrong, then! Right at the start.

Yes, got it!
Thanks a lot @PeroK :)