Integrating ln(x+1)/(x^2+1) from 0 to 1: Need Help!

[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!

 

[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.

I hope that helps. You can check your answer here: http://integrals.wolfram.com/index.jsp

You know, if you'd like.

 

[LCKurtz]

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

 

[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.

 

[rock.freak667]

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.

∫ln(x)dx= xln(x)-x +C

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

this seems most true.

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

 

[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.

 

[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

 

[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?

 

[Bohrok]

Take a look at this thread with the same definite integral:
https://www.physicsforums.com/showthread.php?t=326935

 

[vm310]

Thanks Bohrok that link helped a lot. :D

 

[Count Iblis]

Now try a similar one on your own:

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