Indefinite Integration

\(\int \log x \cdot(\log x+2) d x=\)

\(\int \sin ^{-1} x d x=\)

\(\int \cot x \cdot \log [\log (\sin x)] d x=\)

If \(\int \sqrt{x-\frac{1}{x}}\left(\frac{x^2+1}{x^2}\right) d x=\frac{2}{3}\left(x-\frac{1}{x}\right)^k+c\), then value of \(k\) is

If \(\int \frac{\sin \theta}{\sin 3 \theta} d \theta=\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c\), then \(k=\)

\(\int \frac{d x}{\cos 2 x-\cos ^2 x}=\)

\(\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=\)

\(\int\left[-\frac{\log x-1}{1+(\log x)^2}\right]^2 d x=\)

\(\int \frac{d x}{(\sin x+\cos x)(2 \cos x+\sin x)}=\)

If $\int \frac{1}{1-\cot x} d x=A x+B \log |\sin x-\cos x|+c$ then $A+B=\ldots \ldots$

\(\int \log x \cdot[\log (e x)]^{-2} d x=\ldots\)

\(\int \frac{\sqrt{x^2-a^2}}{x} d x=\ldots \ldots\)

\(\int \frac{\cos x+x \sin x}{x^2+x \cos x} d x=\) ...........

\(\int \frac{1}{\left(x^2+1\right)^2} d x=\ldots\)

\(\int \frac{x^2+1}{x^4-x^2+1} d x=\ldots \ldots\)

$$\begin{aligned} & \text { If } \int \tan (x-\alpha) \tan (x+\alpha) \cdot \tan 2 x d x \\ & =p \log |\sec 2 x|+q \log |\sec (x+\alpha)| \\ & +r \log |\sec (x-\alpha)|+c \text { then } p+q+r=\ldots \ldots \ldots \end{aligned}$$

If \(\int \frac{\cos x-\sin x}{8-\sin 2 x} d x=\frac{1}{p} \log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+c,\) then $p=$ .............