Homework Help: Stuck on this Integral

1. Feb 5, 2010

vm310

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

The limits of integration are from 0 to 1.

I've tried all the ways I've learned how to approach an integral so far and am stuck. (tried u-sub and by parts.)

If anyone could throw me a bone that'd be great.

Thanks!

2. Feb 5, 2010

krtica

Natural log cannot be integrated, but only differentiated.

Right away, you should know to use integration by parts. The quantity ln(x+1) should be chosen for "u", because choosing it for "v" would only put you where you began.

You know, if you'd like.

Last edited: Feb 5, 2010
3. Feb 5, 2010

LCKurtz

I suspect a misprint or mis-copied problem. Apparently there is no elementary antiderivative.

4. Feb 5, 2010

vm310

Thanks for the help guys but i'm still stuck on this problem. I tried inputting it into wolfram online integrator and it didn't really help. I also tried by parts with the u set to what was recommended above.

5. Feb 5, 2010

rock.freak667

∫ln(x)dx= xln(x)-x +C

this seems most true.

vm310, are you 100% sure you copied that problem correctly?

6. Feb 5, 2010

vm310

yeah 100%. It's suppose to be a bonus question with some sort of prize if we can work it out on paper. The problem was copied correctly on the original post, i just forgot to put a dx at the end. when i plug the equation into my calculator i get 0.272198.

7. Feb 5, 2010

vm310

here's what i've got so far:

u=ln(x+1)

du=$$\frac{dx}{(x+1)}$$

dv=$$\frac{dx}{x^2+1}$$

v=arctan(x)

using integration by parts i got:

arctan(x)ln(x+1)-$$\int$$$$\frac{arctan(x)dx}{(x+1)}$$

than using integration by parts again i got:

u=arctan(x)

du=$$\frac{dx}{1+x^2}$$

dv=$$\frac{dx}{x+1}$$

v=ln(x+1)

$$\int$$$$\frac{ln(x+1)dx}{x^2+1}$$ = arctan(x)ln(x+1)-ln(x+1)arctan(x)+$$\int$$$$\frac{ln(x+1)dx}{1+x^2}$$

Help

8. Feb 5, 2010

xcvxcvvc

If you are sure that is the correct integral, perhaps you are leaving out part of the question that would aid us in evaluating the integral without integration. Without any trick like that, this integral is unsolvable. Maybe you're supposed to approximate it and show your work?

9. Feb 6, 2010

Bohrok

10. Feb 6, 2010

vm310

Thanks Bohrok that link helped a lot. :D

11. Feb 6, 2010

Count Iblis

Now try a similar one on your own:

$$\int_{0}^{\pi}\ln\left[\sin(x)\right]dx$$