Embibe Experts Solutions for Chapter: Indefinite Integration, Exercise 1: Exercise 1

What is $\int {\tan }^{-1}\left(\sec x+\tan x\right)dx$ equal to $?$

Let $f\left(x\right)$ and $g\left(x\right)$ be twice differentiable functions on $\left[0,2\right]$ satisfying $f^{\text{'}\text{'}}\left(x\right)=g^{\text{'}\text{'}}\left(x\right), f^{\text{'}}\left(1\right)=4$$,$ $g^{\text{'}}\left(1\right)=6, f\left(2\right)=3$ and $g\left(2\right)=9$. Then, what is $f\left(x\right)-g\left(x\right)$ at $x=4$ equal to ?

$\int \frac{e^x(x+3)}{(x+5{)}^3}dx=$

$\int \frac{\sec x}{\sqrt{\sin (2x+\theta )+\sin \theta }}dx=$

The value of the integral $\int \frac{\sin \theta ·\sin 2\theta \left({\sin }^6\theta +{\sin }^4\theta +{\sin }^2\theta \right)\sqrt{2{\sin }^4\theta +3{\sin }^2\theta +6}}{1-\cos 2\theta }d\theta$ is (where $c$ is a constant of integration)

What is $\int \frac{x{e}^{x}dx}{{\left(x+1\right)}^2}$ equal to ?

$\int \frac{dx}{1+e^{-x}}$ is equal to

$\int {\left(\ln x\right)}^{-1}dx-\int {\left(\ln x\right)}^{-2}dx$ is equal to