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

Let $f^{\text{'}}\left(x\right)=3x^2·\sin \frac{1}{x}-x\cos \frac{1}{x},x\ne 0,f\left(0\right)=0,f\left(\frac{1}{\pi }\right)=0$, then which of the following is/are not correct.

If $\int \left[\frac{\sqrt{x^2+1}\left[\ln \left(x^2+1\right)-2\ln \left(x\right)\right]}{x^4}\right]dx=\frac{\left(x^2+1\right)\sqrt{x^2+1}}{a^2{x}^3}\left(2-a\ln \left(1+\frac{1}{x^2}\right)\right)+c$ then the value of $a$ is

If $f\left(x\right)={\tan }^{-1}x+\ln \sqrt{1+x}-\ln \sqrt{1-x}$ and $\frac{1}{2}\int f^{\text{'}}\left(x\right)d\left(x^4\right)=-\ln \left(1-x^n\right)+C$, then the value of $n$ is

If $\int \sqrt{\frac{cosecx-\cot x}{cosecx+\cot x}}·\frac{\sec x}{\sqrt{1+2\sec x}}dx={\sin }^{-1}\left(\frac{1}{k}{\sec }^2\frac{x}{2}\right)+C$ then the value of $k$ is

If $\int \cos 2\theta ·\ln \left(\frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }\right)d\theta =\frac{1}{\mathrm{k}}\sin 2\theta \left(\mathrm{lntan}\left(\frac{\pi }{4}+\theta \right)\right)-\frac{1}{\mathrm{k}}\ln \left(\sec 2\theta \right)$ then the value of $k$ is

If $\int \frac{x\ln \left(x\right)}{{\left(x^2-1\right)}^{3/2}}dx=f\left(x\right)+C$ and $f\left(\sqrt{2}\right)=\frac{3\pi }{k}-\frac{\ln 2}{2}$, then the value of $k$ is

If $\int {\left(\sin x\right)}^{-11/3}{\left(\cos x\right)}^{-1/3}dx=f\left(x\right)$ and $f\left(\frac{\pi }{2}\right)=0$, then $\left|4f\left(\frac{\pi }{4}\right)\right|$ is

If $\int \frac{2x^2+3x+3}{\sqrt{x^2+2x+2}}dx=ax\sqrt{x^2+2x+2}+b\ln \left|\left(x+1\right)+\sqrt{{\left(x+1\right)}^2+1}\right|+c$ ($c$ is integration constant) then $a+b$ is equal to