Evaluate \[ \int \ln\sqrt{x^2+1}\,dx \] and \[ \int \frac{2\arctan(e^x)}{e^{2x}}\,dx. \]
Solution
Since \(\ln\sqrt{x^2+1}=\tfrac12\ln(x^2+1)\), integration by parts gives \[ \begin{aligned} \int \ln\sqrt{x^2+1}\,dx &=\frac x2\ln(x^2+1)-\int\frac{x^2}{x^2+1}\,dx\\ &=\boxed{\frac x2\ln(x^2+1)-x+\arctan x+C.} \end{aligned} \]
For the second integral, integrate by parts with \[ u=\arctan(e^x),\qquad dv=2e^{-2x}\,dx, \] so \(du=e^x(1+e^{2x})^{-1}dx\) and \(v=-e^{-2x}\). Thus \[ I=-e^{-2x}\arctan(e^x)+\int\frac{e^{-x}}{1+e^{2x}}\,dx. \] Using \[ \frac{e^{-x}}{1+e^{2x}}=e^{-x}-\frac{e^x}{1+e^{2x}}, \] we obtain \[ \boxed{I=-e^{-2x}\arctan(e^x)-e^{-x}-\arctan(e^x)+C.} \]