Solution

The anti derivative of $\sin 2 x$ is a function of $x$ whose derivative is $\sin 2 x$.

It is known that,

$\dfrac{d}{d x}(\cos 2 x)=-2 \sin 2 x$

$\Rightarrow \sin 2 x=-\dfrac{1}{2} \dfrac{d}{d x}(\cos 2 x)$

$\therefore \sin 2 x=\dfrac{d}{d x}(-\dfrac{1}{2} \cos 2 x)$

Therefore, the anti derivative of $\sin 2 x$ is $-\dfrac{1}{2} \cos 2 x$

Solution

The anti derivative of $\cos 3 x$ is a function of $x$ whose derivative is $\cos 3 x$.

It is known that,

$\dfrac{d}{d x}(\sin 3 x)=3 \cos 3 x$

$\Rightarrow \cos 3 x=\dfrac{1}{3} \dfrac{d}{d x}(\sin 3 x)$

$\therefore \cos 3 x=\dfrac{d}{d x}(\dfrac{1}{3} \sin 3 x)$

Therefore, the anti derivative of $\cos 3 x$ is $\dfrac{1}{3} \sin 3 x$.

Solution

The anti derivative of $e^{2 x}$ is the function of $x$ whose derivative is $e^{2 x}$.

It is known that, $\dfrac{d}{d x}(e^{2 x})=2 e^{2 x}$

$\Rightarrow e^{2 x}=\dfrac{1}{2} \dfrac{d}{d x}(e^{2 x})$

$\therefore e^{2 x}=\dfrac{d}{d x}(\dfrac{1}{2} e^{2 x})$

Therefore, the anti derivative of $e^{2 x}$ is $\dfrac{1}{2} e^{2 x}$.

Solution

The anti derivative of $(a x+b)^{2}$ is the function of $x$ whose derivative is $(a x+b)^{2}$.

It is known that,

$\dfrac{d}{d x}(a x+b)^{3}=3 a(a x+b)^{2}$

$\Rightarrow(a x+b)^{2}=\dfrac{1}{3 a} \dfrac{d}{d x}(a x+b)^{3}$

$\therefore(a x+b)^{2}=\dfrac{d}{d x}(\dfrac{1}{3 a}(a x+b)^{3})$

Therefore, the anti derivative of $(a x+b)^{2}$ is $\dfrac{1}{3 a}(a x+b)^{3}$.

Solution

The anti derivative of $(\sin 2 x-4 e^{3 x})$ is the function of $x$ whose derivative is $(\sin 2 x-4 e^{3 x})$.

It is known that, $\dfrac{d}{d x}(-\dfrac{1}{2} \cos 2 x-\dfrac{4}{3} e^{3 x})=\sin 2 x-4 e^{3 x}$

Therefore, the anti derivative of $(\sin 2 x-4 e^{3 x})$ is $(-\dfrac{1}{2} \cos 2 x-\dfrac{4}{3} e^{3 x})$.

Solution

$ \begin{aligned} & \int(4 e^{3 x}+1) d x \\ & =4 \int e^{3 x} d x+\int 1 d x \\ & =4(\dfrac{e^{3 x}}{3})+x+C \\ & =\dfrac{4}{3} e^{3 x}+x+C \end{aligned} $

Solution

$\int x^{2}(1-\dfrac{1}{x^{2}}) d x$

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

$=\int x^{2} d x-\int 1 d x$

$=\dfrac{x^{3}}{3}-x+C$

Solution

$\int(a x^{2}+b x+c) d x$

$=a \int x^{2} d x+b \int x d x+c \int 1 \cdot d x$

$=a(\dfrac{x^{3}}{3})+b(\dfrac{x^{2}}{2})+c x+C$

$=\dfrac{a x^{3}}{3}+\dfrac{b x^{2}}{2}+c x+C$

Solution

$\int(2 x^{2}+e^{x}) d x$

$=2 \int x^{2} d x+\int e^{x} d x$

$=2(\dfrac{x^{3}}{3})+e^{x}+C$

$=\dfrac{2}{3} x^{3}+e^{x}+C$

Solution

$ \begin{aligned} & \int(\sqrt{x}-\dfrac{1}{\sqrt{x}})^{2} d x \\ & =\int(x+\dfrac{1}{x}-2) d x \\ & =\int x d x+\int \dfrac{1}{x} d x-2 \int 1 \cdot d x \\ & =\dfrac{x^{2}}{2}+\log |x|-2 x+C \end{aligned} $

Solution

$ \begin{aligned} & \int \dfrac{x^{3}+5 x^{2}-4}{x^{2}} d x \\ & =\int(x+5-4 x^{-2}) d x \\ & =\int x d x+5 \int 1 \cdot d x-4 \int x^{-2} d x \\ & =\dfrac{x^{2}}{2}+5 x-4(\dfrac{x^{-1}}{-1})+C \\ & =\dfrac{x^{2}}{2}+5 x+\dfrac{4}{x}+C \end{aligned} $

Solution

$ \begin{aligned} & \int \dfrac{x^{3}+3 x+4}{\sqrt{x}} d x \\ & =\int(x^{\dfrac{5}{2}}+3 x^{\dfrac{1}{2}}+4 x^{-\dfrac{1}{2}}) d x \\ & =\dfrac{x^{\dfrac{7}{2}}}{\dfrac{7}{2}}+\dfrac{3(x^{\dfrac{3}{2}})}{\dfrac{3}{2}}+\dfrac{4(x^{\dfrac{1}{2}})}{\dfrac{1}{2}}+C \\ & =\dfrac{2}{7} x^{\dfrac{7}{2}}+2 x^{\dfrac{3}{2}}+8 x^{\dfrac{1}{2}}+C \\ & =\dfrac{2}{7} x^{\dfrac{7}{2}}+2 x^{\dfrac{3}{2}}+8 \sqrt{x}+C \end{aligned} $

Solution

$\int \dfrac{x^{3}-x^{2}+x-1}{x-1} d x$

On dividing, we obtain

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

$=\int x^{2} d x+\int 1 d x$

$=\dfrac{x^{3}}{3}+x+C$

Solution

$\int(1-x) \sqrt{x} d x$

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

$=\int x^{\dfrac{1}{2}} d x-\int x^{\dfrac{3}{2}} d x$

$=\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}}-\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}}+C$

$=\dfrac{2}{3} x^{\dfrac{3}{2}}-\dfrac{2}{5} x^{\dfrac{5}{2}}+C$

Solution

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

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

$=3 \int x^{\dfrac{5}{2}} d x+2 \int x^{\dfrac{3}{2}} d x+3 \int x^{\dfrac{1}{2}} d x$

$=3(\dfrac{x^{\dfrac{7}{2}}}{\dfrac{7}{2}})+2(\dfrac{x^{\dfrac{5}{2}}}{\dfrac{5}{2}})+3 \dfrac{(x^{\dfrac{3}{2}})}{\dfrac{3}{2}}+C$

$=\dfrac{6}{7} x^{\dfrac{7}{2}}+\dfrac{4}{5} x^{\dfrac{5}{2}}+2 x^{\dfrac{3}{2}}+C$

Solution

$\int(2 x-3 \cos x+e^{x}) d x$

$=2 \int x d x-3 \int \cos x d x+\int e^{x} d x$

$=\dfrac{2 x^{2}}{2}-3(\sin x)+e^{x}+C$

$=x^{2}-3 \sin x+e^{x}+C$

Solution

$\int(2 x^{2}-3 \sin x+5 \sqrt{x}) d x$ $=2 \int x^{2} d x-3 \int \sin x d x+5 \int x^{\dfrac{1}{2}} d x$

$=\dfrac{2 x^{3}}{3}-3(-\cos x)+5(\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}})+C$

$=\dfrac{2}{3} x^{3}+3 \cos x+\dfrac{10}{3} x^{\dfrac{3}{2}}+C$

Solution

$\int \sec x(\sec x+\tan x) d x$

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

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

$=\tan x+\sec x+C$

Solution

$\int \dfrac{\sec ^{2} x}{cosec^{2}x } d x$

$ \begin{aligned} & =\int \dfrac{\dfrac{1}{\cos ^{2} x}}{\dfrac{1}{\sin ^{2} x}} d x \\ & =\int \dfrac{\sin ^{2} x}{\cos ^{2} x} d x \\ & =\int \tan ^{2} x d x \\ & =\int(\sec ^{2} x-1) d x \\ & =\int \sec ^{2} x d x-\int 1 d x \\ & =\tan x-x+C \end{aligned} $

Solution

$\int \dfrac{2-3 \sin x}{\cos ^{2} x} d x$

$=\int(\dfrac{2}{\cos ^{2} x}-\dfrac{3 \sin x}{\cos ^{2} x}) d x$

$=\int 2 \sec ^{2} x d x-3 \int \tan x \sec x d x$

$=2 \tan x-3 \sec x+C$

Solution

$ (\sqrt{x}+\dfrac{1}{\sqrt{x}}) d x $

$=\int x^{\dfrac{1}{2}} d x+\int x^{-\dfrac{1}{2}} d x$

$=\dfrac{x^{\dfrac{3}{2}}}{\dfrac{3}{2}}+\dfrac{x^{\dfrac{1}{2}}}{\dfrac{1}{2}}+C$

$=\dfrac{2}{3} x^{\dfrac{3}{2}}+2 x^{\dfrac{1}{2}}+C$

Hence, the correct Answer is C.

Solution

It is given that,

$\dfrac{d}{d x} [f(x)]=4 x^{3}-\dfrac{3}{x^{4}}$

$\therefore$ Anti derivative of $4 x^{3}-\dfrac{3}{x^{4}}=f(x)$

$\therefore f(x)=\int 4 x^{3}-\dfrac{3}{x^{4}} d x$

$f(x)=4 \int x^{3} d x-3 \int(x^{-4}) d x$

$f(x)=4(\dfrac{x^{4}}{4})-3(\dfrac{x^{-3}}{-3})+C$

$\therefore(x)=x^{4}+\dfrac{1}{x^{3}}+C$

Also,

$f(2)=0$

$\therefore f(2)=(2)^{4}+\dfrac{1}{(2)^{3}}+C=0$

$\Rightarrow 16+\dfrac{1}{8}+C=0$

$\Rightarrow C=-(16+\dfrac{1}{8})$

$\Rightarrow C=\dfrac{-129}{8}$

$\therefore f(x)=x^{4}+\dfrac{1}{x^{3}}-\dfrac{129}{8}$

Hence, the correct Answer is A.