Question

Integrate from 0 to 1

tan^-1(x)/(1+x) dx

Answer 1

\[\int\limits_{0}^{1}\frac{ \tan^{-1} x }{ 1+x } dx\]

Answer 2

you sure it is not \[\int\limits_{0}^{1}\frac{ \tan^{-1} x }{ 1+x^2 } dx\]

Answer 3

cause i seriously doubt you are going to find an anti derivative for \[\frac{\tan^{-1}(x)}{1+x}\]

Answer 4

Nope, not 1+x^2

if it were 1+x^2

the final answer will be (tan^(-1)x)^2 / 2 ] from 0 to 1

Answer 5

where did the question come from? you are not going to find a closed form for your anti derivative

Answer 6

.boredofstudies

Answer 7

final answer

pi/4 log2 - pi/8 log 2

Answer 8

which is pi log2/8

Answer 9

wow
how did you do it?

Answer 10

:D wolfram xDDD

Answer 11

gotta love wolfram !

Answer 12

No steps :/

Answer 13

Here's a possible starting point.

Call the integral \(I\). Take \(\arctan x=y\) (so that for \(|x|<\dfrac{\pi}{2}\) we have \(x=\tan y\)). This means \(\mathrm{d}x=\sec^2y\,\mathrm{d}y\). Now,
\[I=\int_0^{\pi/4} \frac{y\sec^2y}{1+\tan y}\,\mathrm{d}y\]
Integrating by parts, we could use
\[\begin{matrix}u=y&&&\mathrm{d}v=\dfrac{\sec^2y}{1+\tan y}\,\mathrm{d}y\\[1ex]
\mathrm{d}u=\mathrm{d}y&&&v=\ln(1+\tan y)\end{matrix}\]so that
\[\begin{align*}I&=\bigg[y\ln(1+\tan y)\bigg]_{y=0}^{y=\pi/4}-\int_0^{\pi/4}\ln(1+\tan y)\,\mathrm{d}y\\[1ex]
&=\frac{\pi\ln2}{4}-\int_0^{\pi/4}\ln(1+\tan y)\,\mathrm{d}y
\end{align*}\]which gives you what you know is at least part of the answer.

An interesting thing to note is that since the final result for the definite integral is expected to be \(\dfrac{\pi\ln2}{8}\), you could try showing that \(I\) is exactly the same as the remaining integral. That is,
\[2I=\frac{\pi\ln2}{4}\implies I=\frac{\pi\ln2}{8}\]
How to do this isn't immediately clear to me.

Another possible route is to consider the series expansion of \(\ln(1+\xi)=\xi-\dfrac{\xi^2}{2}+\dfrac{\xi^3}{3}-\dfrac{\xi^4}{4}+\cdots\) with \(\xi=\tan y\), but I haven't done enough investigating to come up with anything conclusive.

Answer 14

@anonymous
For the remaining integral : use this property \[\int\limits_{0}^{\pi/4} \ln(1+tany) dy = \int\limits_{0}^{\pi/4} \ln(1+ \tan(\frac{ \pi }{ 4 } - y) dy\]

\[\int\limits_{0}^{\pi/4} \ln(\frac{ 1 + tany }{ 1+ tany }+ \frac{ 1 - tany }{ 1+ tany}) dy\]

\[\int\limits_{0}^{\pi/4} \ln(2) - \ln(1 + tany) dy\]
\[2 I_{1} = \int\limits_{0}^{\pi/4} \ln(2) dy = [\ln(2) y ]_{0}^{\pi/4} = \ln(2)\frac{ \pi }{ 4 }\]

\[I_{1} = \frac{ \ln2 \pi }{ 8 }\]

Which adds up with your result to ln(2)pi/8

Nice work mate , keep it up.