Find the indefinite integral of ln(x) * arctan(x) with respect to x.

Step-by-step solution

1. 1

   Choose which factor to differentiate. Both $\ln x$ and $\tan^{-1}x$ have simple derivatives but awkward antiderivatives. Taking $u=\ln x$ (so $du=\frac{dx}{x}$) and $dv=\tan^{-1}(x)\,dx$ works better, because the resulting $\frac{1}{x}$ will cancel against the $x$ inside $v$ — the choice that makes the leftover integral simpler rather than worse.

2. 2

   Find $v=\int\tan^{-1}x\,dx$ by a preliminary integration by parts. With $a=\tan^{-1}x$ and $db=dx$:

   $\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^{2}}dx=x\tan^{-1}x-\tfrac12\ln(1+x^{2}).$

   The last integral is a $u$-substitution with $u=1+x^{2}$, giving the $\tfrac12$.

3. 3

   Apply the main integration by parts. With $u=\ln x$ and $v=x\tan^{-1}x-\tfrac12\ln(1+x^{2})$:

   $\int\ln x\,\tan^{-1}x\,dx=\ln x\left(x\tan^{-1}x-\tfrac12\ln(1+x^{2})\right)-\int\left(x\tan^{-1}x-\tfrac12\ln(1+x^{2})\right)\frac{dx}{x}.$

   Dividing through by $x$ inside the integral gives

   $\int\ln x\,\tan^{-1}x\,dx=x\ln x\tan^{-1}x-\tfrac12\ln x\ln(1+x^{2})-\int\tan^{-1}x\,dx+\tfrac12\int\frac{\ln(1+x^{2})}{x}dx.$

   This is exactly the cancellation predicted in step 1: $x\tan^{-1}x\cdot\frac1x$ collapses to $\tan^{-1}x$, an integral already computed.

4. 4

   Recognise the remaining integral as a dilogarithm. The piece $\int\frac{\ln(1+x^{2})}{x}dx$ has no elementary antiderivative. Substituting $t=-x^{2}$ and comparing with the definition $\operatorname{Li}_{2}(z)=-\int_{0}^{z}\frac{\ln(1-w)}{w}dw$ gives

   $\int\frac{\ln(1+x^{2})}{x}dx=-\tfrac12\operatorname{Li}_{2}(-x^{2})+C.$

   A numeric check: differentiating $-\tfrac12\operatorname{Li}_{2}(-x^{2})$ at $x=0.6$ gives $0.5124745$, and $\frac{\ln(1.36)}{0.6}=0.5124745$ ✓.

5. 5

   Assemble the closed form. Substituting $\int\tan^{-1}x\,dx$ from step 2 and the dilogarithm from step 4:

   $\int\ln x\,\tan^{-1}x\,dx=x\ln x\tan^{-1}x-\tfrac12\ln x\ln(1+x^{2})-x\tan^{-1}x+\tfrac12\ln(1+x^{2})-\tfrac14\operatorname{Li}_{2}(-x^{2})+C.$

   So the answer exists in closed form, but only once the dilogarithm is admitted — no combination of elementary functions will do.

6. 6

   Verify by differentiating numerically. Evaluating the derivative of the right-hand side with a central difference: at $x=0.5$ it gives $-0.32137603$ against $\ln(0.5)\tan^{-1}(0.5)=-0.32137603$ ✓; at $x=0.7$, $-0.21783065$ against $-0.21783065$ ✓. Agreement to eight digits at two independent points confirms every coefficient.