$\frac{4 x^2+5}{(x-2)^4}=\frac{A}{(x-2)}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)^3}+\frac{D}{(x-2)^4}$, then $\sqrt{\frac{A}{C}+\frac{B}{C}+\frac{D}{C}}$ is equal to

$$ \text { If } \frac{13 x+43}{2 x^2+17 x+30}=\frac{A}{2 x+5}+\frac{B}{x+6} \text {, then } A+B \text { is equal to } $$

If $\int \log \left(6 \sin ^2 x+17 \sin x+12\right) \cos x d x=f(x)+c$, then $f\left(\frac{\pi}{2}\right)$ is equal to

$$ \int \frac{x^3 \tan ^{-1} x^4}{1+x^8} d x= $$

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

$$ \int \frac{\sin 7 x}{\sin 2 x \sin 5 x} d x= $$

If $\frac{1}{(3 x+1)(x-2)}=\frac{A}{3 x+1}+\frac{B}{x-2}$ and $\frac{x+1}{(3 x+1)(x-2)}=\frac{C}{3 x+1}+\frac{D}{x-2}$, then

$$ \begin{aligned} &\text { If } \int \frac{3}{2 \cos ^3 x \sqrt{2 \sin 2 x}} d x=\frac{3}{2}(\tan x)^B+\frac{3}{10}(\tan x)^A+C \text {, than }\\&A= \end{aligned} $$

$$\frac{2 x^2+1}{x^3-1}=\frac{A}{x-1}+\frac{B x+C}{x^2+x+1} \Rightarrow 7 A+2 B+C=$$

$$\int \frac{3 x+4}{x^3-2 x+4} d x=\log f(x)+C \Rightarrow f(3)=$$

$$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x=$$

$$\int \frac{d x}{(x-3)^{\frac{4}{5}}(x+1)^{\frac{6}{5}}}=$$

If $$I_n=\int\left(\cos ^n x+\sin ^n x\right) d x$$ and $$I_n-\frac{n-1}{n} I_{n-2} =\frac{\sin x \cos x}{n} f(x)$$, then $$f(x)=$$

If $$f(x)=\int x^2 \cos ^2 x\left(2 x \tan ^2 x-2 x-6 \tan x\right) d x$$ and $$f(0)=\pi$$, then $$f(x)=$$

If $$\int \frac{e^{\sqrt{x}}}{\sqrt{x}}(x+\sqrt{x}) d x=e^{\sqrt{x}}[A x+B \sqrt{x}+C]+K$$ then $$A+B+C=$$

If $$\int \frac{1+\sqrt{\tan x}}{\sin 2 x} d x=A \log \tan x+B \tan x+C$$, then $$4 A-2 B=$$

$$\int \frac{1+\tan x \tan (x+a)}{\tan x \tan (x+a)} d x=$$

Assertion (A) If $$I_n=\int \cot ^n x d x$$, then $$I_6+I_4=\frac{-\cot ^5 x}{5}$$

Reason (R) $$\int \cot ^n x d x=\frac{-\cot ^{n-1} x}{n} -\int \cot ^{n-2} x d x$$

If $$I_n=\int \tan ^n x d x$$, and $$I_0+I_1+2 I_2+2 I_3+2 I_4 +I_5+I_6=\sum_\limits{k=1}^n \frac{\tan ^k x}{k}$$, then $$n=$$

$$\int \frac{e^{\cot x}}{\sin ^2 x}(2 \log \operatorname{cosec} x+\sin 2 x) d x=$$

The parametric form of a curve is $$x=\frac{t^3}{t^2-1} y=\frac{t}{t^2-1}$$, then $$\int \frac{d x}{x-3 y}=$$

Given, $$\frac{3 x-2}{(x+1)^2(x+3)}=\frac{A}{x+1} +\frac{B}{(x+1)^2}+\frac{C}{x+3}$$, then $$4 A+2 B+4 C$$ is equal to

$$\int \frac{\sin \alpha}{\sqrt{1+\cos \alpha}} d \alpha$$ is equal to

If $$\int \frac{\cos 4 x+1}{\cot x-\tan x}=k \cos 4 x+C$$, then $$k$$ is equal to

If $$\int\left[\cos (x) \cdot \frac{d}{d x}(\operatorname{cosec}(x)] d x=f(x)+g(x)+c\right.$$ then $$f(x) \cdot g(x)$$ is equal to

If $$\int \frac{(2 x+1)^6}{(3 x+2)^8} d x=P\left(\frac{2 x+1}{3 x+2}\right)^Q+R$$, then $$\frac{P}{Q}$$ is equal to

Which of the following is partial fraction of $$\frac{-x^2+6 x+13}{(3 x+5)\left(x^2+4 x+4\right)}$$ is equal to

$$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}} d x$$ is equal to

$$\int(\cos x) \log \cot \left(\frac{x}{2}\right) d x$$ is equal to

$$\int \sqrt{e^{4 x}+e^{2 x}} d x$$ is equal to

If $$\int \frac{1}{1+\sin x} d x=\tan (f(x))+c$$, then $$f^{\prime}(0)$$ is equal to

$$\int \frac{e^x(x+3)}{(x+5)^3} d x$$ is equal to

If $$\int \frac{(x-1)^2}{\left(x^2+1\right)^2} d x=\tan ^{-1}(x)+g(x)+k$$, then $$g(x)$$ is equal to

If $$\int \frac{1-(\cot x)^{2021}}{\tan x+(\cot x)^{2022}} d x=\frac{1}{A} \log\left|(\sin x)^{2023}+(\cos x)^{2023}\right|+c$$, then $$A$$ is equal to