Indefinite Integration · Mathematics · TS EAMCET

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

$$ \begin{aligned} &\text { If } y=f(x)^{g(x)} \text { and } \frac{d y}{d x}=y\left[H(x) f^{\prime}(x)+G(x) g^{\prime}(x)\right] \text {, then }\\ &\int \frac{G(x) H(x) f^{\prime}(x)}{g(x)} d x= \end{aligned} $$

$$ \begin{aligned} I_1 & =\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, I_2 \\ & =\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x, \text { then } I_2-I_1= \end{aligned} $$

If $\int \frac{1}{x} \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} d x=2 f(x)-2 \sin ^{-1} \sqrt{x}+c$, then $f(x)=$

$$ \begin{aligned} & \int \frac{3 x+2}{4 x^2+4 x+5} d x=A \log \\ & \left(4 x^2+4 x+5\right)+B \tan ^{-1}\left(\frac{2 x+1}{2}\right)+C, \text { then } A+B= \end{aligned} $$

If $\int x^3 \sin 3 x d x=\frac{1}{27}[f(x) \cos 3 x+g(x) \sin 3 x]+C$, then $f(\mathrm{l})+g(\mathrm{l})=$

If $I_1=\int \sin ^6 x d x$ and $I_2=\int \cos ^6 x d x$, then $I_1+I_2=$

$$ \int \frac{x+\cos x}{1-\sin x} d x= $$

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

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

$$ \int \frac{3^x(x \log 3-1)}{x^2} d x= $$

If $\frac{5 \pi}{4} < x < \frac{7 \pi}{4}$, then $\int \sqrt{\frac{1-\sin 2 x}{1+\sin 2 x}} d x=$

$$ \int x \tan ^{-1} \sqrt{\frac{1+x^2}{1-x^2}} d x= $$

$$ \int \frac{1}{(2 \cos x+\sin x)^2} d x= $$

If $\frac{x+3}{(x+1)\left(x^2+2\right)}=\frac{a}{x+1}+\frac{b x+c}{x^2+2}$, then $a-b+c=$

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

$$ \int\left(1+\tan ^2 x\right)(1+2 x \tan x) d x= $$

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

$$ \int \frac{\log x}{(1+x)^3} d x= $$

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

$$ \int\left(\frac{1}{x^2}+\frac{\sin ^3 x+\cos ^3 x}{\sin ^2 x \cos ^2 x}\right) d x= $$

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

$\int \frac{x^3}{x^4+3 x^2+2} d x=$

If $\int \frac{d x}{\left(x^2+9\right) \sqrt{x^2+16}}=\frac{1}{3 \sqrt{7}} \tan ^{-1}\left(K \frac{x}{\sqrt{16+x^2}}\right)+c$, then $K=$

$$ \int \frac{2 \sin x-3 \cos x}{4 \cos x-3 \sin x} d x= $$

$$ \int e^{4 x}(\sin 3 x-\cos 3 x) d x= $$

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

If $\int(x+2) \sqrt{x^2-x+2} d x=\frac{1}{3} f(x)+\frac{5}{8} g(x)+\frac{35}{16} h(x)+C$ then $f(-1)+g(-1)+h\left(\frac{1}{2}\right)=$

If $\int \frac{1}{x^{4}+8 x^{2}+9} d x=\frac{1}{k}$$\left[\frac{1}{\sqrt{14}} \tan ^{-1}(f(x))-\frac{1}{\sqrt{2}} \tan ^{-1}(g(x))\right]+c$ then,

$\sqrt{\frac{k}{2}+f(\sqrt{3})+g(1)}=$

$\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x=f(x)+c$, where $c$ is the constant of integration. If $\frac{5 \pi}{2}$<$x<\frac{7 \pi}{2}$ and $$ f\left(\frac{8 \pi}{3}\right)=-2, \text { then } f^{\prime}\left(\frac{8 \pi}{3}\right)= $$

$$ \int \frac{\tan x}{\sec ^2 x\left(1+\sec ^6 x\right)^{\frac{2}{3}}} d x= $$

$$ \int \frac{1}{(x-1)^{\frac{5}{7}}(x+1)^{\frac{9}{7}}} d x= $$

$\int \frac{1+\sqrt{3} \cot x}{1-\sqrt{3} \cot x} d x=$

$$ \begin{aligned} & \text { If } \int \frac{1}{\operatorname{cosec} x+\cos x} d x=\frac{1}{2 \sqrt{3}} \log |f(x)| \\ & -\int \frac{\cos x-\sin x}{2+\sin 2 x} d x+c, \text { then at } x=\frac{\pi}{3},|f(x)|= \end{aligned} $$

If $\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^2+C$, then $k=$

If $\int(\log x)^3 x^5 d x=\frac{x^6}{A}\left[B(\log x)^3\right. \left.+C(\log x)^2+D(\log x)-1\right]+k$ and $A, B, C, D$ are integers, then $A-(B+C+D)=$

$$ \int \frac{d x}{\left(x^2+1\right)\left(x^2+4\right)}= $$

$\int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}}=$

$\int \frac{1}{\left(x+\frac{2}{x}\right) \sqrt{x^4+4 x^2+3}} d x=$

If $\frac{3 \pi}{2} < x < \frac{5 \pi}{2}$ and $\int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x=f(x)+C$, where $C$ is the constant of integration, then $f\left(\frac{\pi}{3}\right)-f(0)=$

If $\int \frac{2 \sin 2 x-3 \cos x}{2 \sin ^2 x-3 \sin x+4} d x=f(x)+C$, where $C$ is the constant of integration, then $f\left(\frac{\pi}{2}\right)-f(0)=$

$\int \frac{2 x+3}{\sqrt{3 x^2-2 x+1}} d x=$

$$ \int \frac{1}{16-7 \sin ^2 x} d x= $$

$$ \int \frac{\sec ^2 x}{(\sec x+\tan x)^2} d x= $$

$$ \int \frac{1}{3 \cos x-4 \sin x+5} d x= $$

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

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

If $\int x^4(\log x)^3 d x=x^5\left[A(\log x)^3\right]$ $\left.+B(\log x)^2+C \log x+D\right]+k$, then $A+B+C+5 D=$

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

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

If $f(x)=\int \frac{2-3 \sin ^2 x}{1+\cos 2 x} d x$ and $f\left(\frac{\pi}{4}\right)=1$, then $f(0)=$

If $x \neq(2 n+1) \frac{\pi}{2}, n \in Z$ and $\cos x \neq \frac{-1}{2}$, then

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

Given that $\int \frac{1}{x^2+a^2} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+c$.

$$ \begin{aligned} & \text { If } \int \frac{1}{x^4+3 x^2+1} d x=a \tan ^{-1}\left(\frac{b\left(x^2-1\right)}{x}\right) \\ & +c \tan ^{-1}\left(\frac{d\left(x^2+1\right)}{x}\right)+k \end{aligned} $$

where $k$ is a constant of integration, then $5(c+d+a b)=$

If $\frac{2 x^2-3 x+5}{(x-7)^3}=\frac{A}{x-7}+\frac{B}{(x-7)^2}+\frac{C}{(x-7)^3}$, then $2 A-3 B+C=$

If $\frac{3 x^2+a x+3}{(2 x+3)\left(x^2+2\right)}=\frac{3}{2 x+3}+\frac{B x+C}{x^2+2}$, then $a(B+C)=$

If $\frac{3 \pi}{4}

If $\tan \alpha=\frac{4}{3}$, then $\int \frac{1}{3 \cos x-4 \sin x} d x=$

If $x \neq(2 n+1) \frac{\pi}{2}$, then $\int \frac{\cos ^3 x}{(1+\sin x)^4} d x=$

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

Let $g(x)$ be the anti-derivative of $f(x)$. Then, the function for which $\log _e\left(1+(g(x))^2\right)+c$ is an anti-derivative is

If $f(x)=\int\left[\tan ^2 x+\cot ^2 x+\frac{4\left(\sin ^3 x+\cos ^3 x\right)}{\sin ^2 2 x}\right] d x$ and $f\left(\frac{\pi}{4}\right)=0$, then $3\left[f\left(\frac{\pi}{6}\right)+2\right]=$

$\int \sqrt{4 \cos ^2 x-5 \sin ^2 x} \cos x d x=$

If $\frac{42-13 x}{x^2+x-6}=\frac{A}{l x+m}+\frac{B}{p x+q}$, where $l m>0$ and $p q<0$, then $\frac{A l p}{B m q}=$

Given that $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$ and $\frac{d}{d x}\left(\sin h^{-1} x\right)=\frac{1}{\sqrt{1+x^2}}$. Then, $\int \frac{3 x^6-2 x^4+x^2-2}{x^2+1} d x=$

$$ \int \frac{\sin x \cdot \sec ^2 x-\tan x \cdot \sin x+\cos x}{(1-\cos 2 x)} d x= $$

If $f(x)=\int \frac{16 x^7+5 x^{10}}{\left(x^3+2+3 x^8\right)^2} d x(x \geq 0)$ and $f(0)=1$, then the value of $f(-1)$ is

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

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

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 $\int e^x \cos x d x=\frac{e^x}{2}(\cos x+\sin x)$ 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 $g(1)=$

If $\frac{x^2-3 x+2}{(x-4)(x-3)^2}=\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{(x-3)^2}$ then $A+B+C=$

If $\frac{x^2+3}{\left(x^2+1\right)\left(x^2+2\right)}=\frac{A x+B}{x^2+1}+\frac{C x+D}{x^2+2}$ then $A+B+C+D=$

Let $f(x)=\int \frac{2 x^3-3 x^2+4 x-5}{x^2} d x$ and $f(1)=1$. Then, $f(5)=$

If $x>0$ and $x \neq(2 n+1) \frac{\pi}{2}$, then $\int\left(x \sqrt{x}-e^{\log (\sec x \tan x)}+\frac{3 x^2-2 x+1}{x^2}\right) d x=$

$$ \int(2 x-3) \sqrt{3 x+2} d x= $$

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 $(A, f(-1))=$

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

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

If $5(f(x))^2=x f(x)+30$ and

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

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

$$ f(x)+g(x)= $$

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

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

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 $B-2 A+C=$

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

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

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

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

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

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

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

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

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

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

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 $B+2(D+F+E)-C \cdot A=$

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

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

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

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 $\frac{m-l}{r}=$