Indefinite Integration

$$ \begin{aligned} & \text { If } \int \frac{(x+3)}{(x-1)^2(2 x-1)} d x \\ & =\frac{A}{x-1}+B \log (2 x-1)+C \log (x-1)+k, \text { then } A+B+C= \end{aligned} $$

Let $x \neq \frac{-3}{5}, \frac{2}{5}$, if $f\left(\frac{2 x+1}{5 x+3}\right)=x+2$, then $\int f(x) d x=$

If $\int \frac{1+\cos 8 x}{\tan 2 x-\cot 2 x} d x=f(x) \cdot \cos (g(x))+c$, then $f\left(\frac{1}{4}\right)+g\left(\frac{1}{4}\right)=$

If $\int e^x \cos x d x=\frac{e^x}{2}(\cos x+\sin x)$ and
\(\int \frac{\cos \left(\log \left(\frac{2 x+3}{3-2 x}\right)\right)}{(3-2 x)^2} d x=\frac{f(x)}{24}[\cos (g(x))+\sin (g(x))]+c\)
then $g(1)=$

\(\int\left[\frac{x^4-x}{x^{20}}\right]^{1 / 4} d x=\)

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

If $\int x[\log (1+x)]^3 d x=\frac{(1+x)^2}{16}(f(x))+(1+x)(g(x))$, then
\(f(x)+g(x)=\)

If $5(f(x))^2=x f(x)+30$ and
$$ \begin{aligned} & \int \frac{\left(3 x^3+\left(1-30 x^2\right) f(x)\right)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x \\ & =\frac{A}{B x^3+D f(x)}+C \text { then } A+B+D= \end{aligned} $$

\(\int \frac{d x}{(x-2) \sqrt{x^2-3 x+5}}=\)

If $\int e^x\left(\frac{x^2-8 x+19}{(x-1)^5}\right) d x=\frac{e^x(l x+m)}{(x-1)^4}+C$, then $4 l+m=$

If $f\left(\frac{2 x+3}{3 x+5}\right)=x+4, x \neq \frac{-5}{3}, \frac{2}{3}$ and $\int f(x) d x=A x+B \ln |3 x-2|+C$, then $3 B-A=$

If $\int \frac{(x-1) d x}{(x+1) \sqrt{x^3+x^2+x}}=A \cdot \tan ^{-1} \sqrt{f(x)}+$ constant, then the ordered pair $(A, f(-1))=$

\(I_{m, n}=\int x^m(\log x)^n d x=\)

\(\int \frac{25 x^2+8}{\sqrt{25 x^2+9}} d x=\)

If $\int e^{\sin ^2 x}\left(\sin x \cos x+\cos ^3 x \sin x\right) d x=e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

Let $I_n=\int \sec ^n x d x$. If $5 I_6-4 I_4=f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

If $I_m=\int x^m \cos n x d x=g(x)-\frac{m(m-1)}{n^2} I_{m-2}$, then $g(x)=$

\(\int \frac{d x}{x \ln (x) \ln ^2(x) \ln ^3(x) \ldots \ln ^m(x)}=\frac{(\ln (x))^K}{K}+C \Rightarrow 2 K=\)

\(\int \frac{x^2}{\left(\sqrt{4-x^2}\right)^3} d x=\)

If the partial fractions decomposition of $\frac{x^4+24 x^2+28}{\left(x^2+1\right)^3}$ is $\frac{A}{x^2+1}+\frac{B}{\left(x^2+1\right)^2}+\frac{C}{\left(x^2+1\right)^3}$ then $B-2 A+C=$

If $\int \frac{2 x^{12}+5 x^9}{\left(1+x^3+x^5\right)^3} d x=\frac{x^m}{l\left(1+x^3+x^5\right)^r}+C$, then $\frac{m-l}{r}=$

\(\int e^{-3 x}\left(x^2+\sin 4 x\right) d x=\)

For $k \in(1, \infty), \int \frac{1}{1+k \cos x} d x=$

\(\int \frac{y^2+\sqrt[3]{y^4}+\sqrt[6]{y^2}}{y\left(1+\sqrt[3]{y^2}\right)} d y=\)

If $\frac{4 x^2+5 x^4+7}{\left(x^2+1\right)\left(x^4+x^2+1\right)}=\frac{A x+B}{x^2+1} +\frac{C x^3+D x^2+E x+F}{x^4+x^2+1}$, then $B+2(D+F+E)-C \cdot A=$

$$ \begin{aligned} & \text { If } \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x=A \log (|x \sin x+\cos x|) \\ & +B \frac{f(x)}{(x \tan x+1)}+C \text {, then } f(A+B)= \end{aligned} $$

$$ \begin{aligned} & \text { If } \int \frac{9 x+15}{x^3-6 x-9} d x=A \log |g(x)| \\ & \quad+B \log |f(x)|+C, \text { then } \frac{(A-B) g(4)}{f(-1)}= \end{aligned} $$

$$ \text { If } \begin{aligned} & \int x^3(\log x)^2 d x=x^4\left[A(\log x)^2+B(\log x)\right. \\ &+C \log e]+K, \text { then } A+B+C \end{aligned} $$

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

For $x \in\left(\frac{3 \pi}{4}, \pi\right), \int(\sqrt{1+\sin 2 x}+\sqrt{1-\sin 2 x}) d x=$