Evaluating the indefinite integral $\int\log\left(\tan^{-1}(x)\right)\;\mathrm dx$

Question

I have tried to solve this and have solved it. But I can't.

$$\int\log\left(\tan^{-1}(x)\right)\;\mathrm dx $$

I got this answer $$ \log\left(\tan^{-1}(x)\right)\cdot x. $$ But its wrong. Please help me someone.

Answer 1

Using the substitution $\tan(u)=x$, we get $$ \begin{align} \int\log(\tan^{-1}(x))\,\mathrm{d}x &=\int\log(u)\,\mathrm{d}\tan(u)\\ &=\log(u)\tan(u)-\int\frac{\tan(u)}{u}\mathrm{d}u \end{align} $$ However, I do not believe that $\frac{\tan(u)}{u}$ has an elementary anti-derivative.

Answer 2

$ \int \log(\tan^{-1}(x))\,\mathrm{d}x$ Let $\tan^{-1}(x) = t \Rightarrow \frac{1}{1+x^2}\mathrm{d}x = \mathrm{d}t $ .....(i)

Also x =\tan(t) .......(ii)

Therefore ( by using (ii)): $$ \begin{align} \int\log(\tan^{-1}(x))\,\mathrm{d}x &= \int\log(t) . (1+\tan^2(t))\,\mathrm{d}t\\ &=\int (\log(t) + \log(t) \tan^2(t))dt\\ &= \int( \log(t) + \log(t) \sec^2(t) - \log(t) )\,\mathrm{d}t\\ &= \int \log(t) \sec^2(t)\,\mathrm{d}t \end{align} $$ Which by using integration by parts can easily be solved and using (ILATE ) which means take Inverse function, L-logarithmic function, A-algebric functin, T- Trigonometric function, E - exponential function .

Which can be written as : $$ \log(t) \int \sec^2(t)\,\mathrm{d}t -\int \frac{1}{t}\tan(t)\,\mathrm{d}t = \log(t)\tan(t) -\int \frac{1}{t}\tan(t)\,\mathrm{d}t $$ from here you can easily solve this....