Question

$\int \frac{1+\tan x}{x+\log \sec x}dx$

Answer 1

$$\begin{gathered} \mathrm{Note}:\mathrm{Here},\mathrm{we}\mathrm{are}\mathrm{considering}\log \mathit{}x\mathrm{as}{\log }_{\mathrm{e}}x \\ \mathrm{Let}I=\int \frac{1+\tan x}{x+\log \sec x}dx \\ \mathrm{Putting}x+\log \sec x=t \\ \Rightarrow 1+\frac{\sec x\tan x}{\sec x}=\frac{dt}{dx} \\ \Rightarrow \left(1+\tan x\right)dx=dt \\ \therefore I=\int \frac{1}{t}dt \\ =\log \left|t\right|+C \\ =\log \left|x+\log \sec x\right|+C\left[\because t=x+\log \sec x\right] \end{gathered}$$