Evaluate the five antiderivatives \[ \int\frac{\sqrt{1+\ln x}}{x\ln x}\,dx, \quad \int\frac{7-\ln x}{x(3+\ln x)}\,dx, \] \[ \int\frac{\sec^2x}{1+\tan x}\,dx, \quad \int\frac{\sin x}{\sqrt{4-\cos^2x}}\,dx, \quad \int\frac{\sec^2x}{\sqrt{9-\tan^2x}}\,dx, \] on intervals where the displayed expressions are real and defined.
Solution
For the first, put \(u=\sqrt{1+\ln x}\). Then \(\ln x=u^2-1\) and \(dx/x=2u\,du\), so \[ \frac{\sqrt{1+\ln x}}{x\ln x}\,dx =\frac{2u^2}{u^2-1}\,du =\left(2+\frac1{u-1}-\frac1{u+1}\right)du. \] Hence \[ \boxed{2\sqrt{1+\ln x}+\ln\left|\frac{\sqrt{1+\ln x}-1}{\sqrt{1+\ln x}+1}\right|+C.} \]
For the second, \[ \frac{7-\ln x}{3+\ln x}=-1+\frac{10}{3+\ln x}, \] so \[ \boxed{-\ln x+10\ln|3+\ln x|+C.} \]
For the third, \(u=1+\tan x\) gives \[ \boxed{\ln|1+\tan x|+C.} \]
For the fourth, \(u=\cos x/2\) gives \(du=-\tfrac12\sin x\,dx\) and \[ \boxed{-\arcsin\!\left(\frac{\cos x}{2}\right)+C.} \]
For the fifth, \(u=\tan x/3\) gives \(du=\tfrac13\sec^2x\,dx\) and \[ \boxed{\arcsin\!\left(\frac{\tan x}{3}\right)+C.} \]