Solution

Let $I=\int x \sin x d x$

Taking $x$ as first function and $\sin x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x \int \sin x d x - \int {\dfrac{d}{d x} (x) \int \sin x d x} d x \\ & =x(-\cos x)-\int 1 \cdot(-\cos x) d x \\ & =-x \cos x+\sin x+C \end{aligned} $

Solution

Let $I=\int x \sin 3 x d x$

Taking $x$ as first function and $\sin 3 x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x \int \sin 3 x d x-\int\ {(\dfrac{d}{d x} x) \int \sin 3 x d x}dx \\ & =x(\dfrac{-\cos 3 x}{3})-\int 1 \cdot(\dfrac{-\cos 3 x}{3}) d x \\ & =\dfrac{-x \cos 3 x}{3}+\dfrac{1}{3} \int \cos 3 x d x \\ & =\dfrac{-x \cos 3 x}{3}+\dfrac{1}{9} \sin 3 x+C \end{aligned} $

Solution

$x^{2} e^{x}$

Let $I=\int x^{2} e^{x} d x$

Taking $x^{2}$ as first function and $e^{x}$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =x^{2} \int e^{x} d x-\int {(\dfrac{d}{d x} x^{2}) \int e^{x} d x} d x \\ & =x^{2} e^{x}-\int 2 x \cdot e^{x} d x \\ & =x^{2} e^{x}-2 \int x \cdot e^{x} d x \end{aligned} $

Again integrating by parts, we obtain

$ \begin{aligned} & =x^{2} e^{x}-2[x \cdot \int e^{x} d x-\int{(\dfrac{d}{d x} x) \cdot \int e^{x} d x} d x] \\ & =x^{2} e^{x}-2[x e^{x}-\int e^{x} d x] \\ & =x^{2} e^{x}-2[x e^{x}-e^{x}] \\ & =x^{2} e^{x}-2 x e^{x}+2 e^{x}+C \\ & =e^{x}(x^{2}-2 x+2)+C \end{aligned} $

Solution

Let $I=\int x \log x d x$

Taking $\log x$ as first function and $x$ as second function and integrating by parts, we obtain

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

Solution

Let $I=\int x \log 2 x d x$

Taking $\log 2 x$ as first function and $x$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\log 2 x \int x d x-\int{\dfrac{d}{d x} ( \log2 x) \int x d x} d x \\ & =\log 2 x \cdot \dfrac{x^{2}}{2}-\int \dfrac{2}{2 x} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \log 2 x}{2}-\int \dfrac{x}{2} d x \\ & =\dfrac{x^{2} \log 2 x}{2}-\dfrac{x^{2}}{4}+C \end{aligned} $

Solution

Let $I=\int x^{2} \log x d x$

Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\log x \int x^{2} d x-\int{(\dfrac{d}{d x} \log x) \int x^{2} d x} d x \\ & =\log x(\dfrac{x^{3}}{3})-\int \dfrac{1}{x} \cdot \dfrac{x^{3}}{3} d x \\ & =\dfrac{x^{3} \log x}{3}-\int \dfrac{x^{2}}{3} d x \\ & =\dfrac{x^{3} \log x}{3}-\dfrac{x^{3}}{9}+C \end{aligned} $

Solution

Let $I=\int x \sin ^{-1} x d x$

Taking $\sin ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

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

Solution

Let $I=\int x \tan ^{-1} x d x$

Taking $\tan ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

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

Solution

Let $I=\int x \cos ^{-1} x d x$

Taking $\cos ^{-1} x$ as first function and $x$ as second function and integrating by parts, we obtain

$$ \begin{align*} I & =\cos ^{-1} x \int x d x-\int{\dfrac{d}{d x} (\cos ^{-1} x) \int x d x} d x \\ & =\cos ^{-1} x \dfrac{x^{2}}{2}-\int \dfrac{-1}{\sqrt{1-x^{2}}} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int \dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int\ {\sqrt{1-x^{2}}+(\dfrac{-1}{\sqrt{1-x^{2}}})\} d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} \int \sqrt{1-x^{2}} d x-\dfrac{1}{2} \int(\dfrac{-1}{\sqrt{1-x^{2}}}) d x \\ & =\dfrac{x^{2} \cos ^{-1} x}{2}-\dfrac{1}{2} I_1-\dfrac{1}{2} \cos ^{-1} x \qquad …(1) \end{align*} $$

where, $I_1=\int \sqrt{1-x^{2}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int {\dfrac{d}{d x} (\sqrt{1-x^{2})} \int x d x}dx$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{-2 x}{2 \sqrt{1-x^{2}}} \cdot x d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{-x^{2}}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-\int \dfrac{1-x^{2}-1}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-{\int \sqrt{1-x^{2}} d x+\int \dfrac{-d x}{\sqrt{1-x^{2}}}}$

$\Rightarrow I_1=x \sqrt{1-x^{2}}-{I_1+\cos ^{-1} x}$

$\Rightarrow 2 I_1=x \sqrt{1-x^{2}}-\cos ^{-1} x$

$\therefore I_1=\dfrac{x}{2} \sqrt{1-x^{2}}-\dfrac{1}{2} \cos ^{-1} x$

Substituting in (1), we obtain

$ \begin{aligned} I & =\dfrac{x \cos ^{-1} x}{2}-\dfrac{1}{2}(\dfrac{x}{2} \sqrt{1-x^{2}}-\dfrac{1}{2} \cos ^{-1} x)-\dfrac{1}{2} \cos ^{-1} x \\ & =\dfrac{(2 x^{2}-1)}{4} \cos ^{-1} x-\dfrac{x}{4} \sqrt{1-x^{2}}+C \end{aligned} $

Solution

Let $I=\int(\sin ^{-1} x)^{2} \cdot 1 d x$

Taking $(\sin ^{-1} x)^{2}$ as first function and 1 as second function and integrating by parts, we obtain

$ \begin{aligned} I & =(\sin ^{-1} x)^2 \int 1 d x-\int{\dfrac{d}{d x}(\sin ^{-1} x)^{2} \cdot \int 1 \cdot d x} d x \\ & =(\sin ^{-1} x)^{2} \cdot x-\int \dfrac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x \\ & =x(\sin ^{-1} x)^{2}+\int \sin ^{-1} x \cdot(\dfrac{-2 x}{\sqrt{1-x^{2}}}) d x \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x-\int\ {(\dfrac{d}{d x} \sin ^{-1} x) \int (\dfrac{-2 x}{\sqrt{1-x^{2}}} d x)\} d x] \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \dfrac{1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-\int 2 d x \\ & =x(\sin ^{-1} x)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+C \end{aligned} $

Solution

Let $I=\int \dfrac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$

$\Rightarrow I=\dfrac{-1}{2} \int \dfrac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$

Taking $\cos ^{-1} x$ as first function and $(\dfrac{-2 x}{\sqrt{1-x^{2}}})$ as second function and integrating by parts, we obtain

$ \begin{aligned} I & =\dfrac{-1}{2}[\cos ^{-1} x \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x-\int{\dfrac{d}{d x} (\cos ^{-1} x) \int \dfrac{-2 x}{\sqrt{1-x^{2}}} d x} d x] \\ & =\dfrac{-1}{2}[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \dfrac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x] \\ & =\dfrac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x] \\ & =\dfrac{-1}{2}[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x]+C \\ & =-[\sqrt{1-x^{2}} \cos ^{-1} x+x]+C \end{aligned} $

Solution

Let $I=\int x \sec ^{2} x d x$

Taking $x$ as first function and $\sec ^{2} x$ as second function and integrating by parts, we obtain

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

Solution

Let $I=\int 1 \cdot \tan ^{-1} x d x$

Taking $\tan ^{-1} x$ as first function and 1 as second function and integrating by parts, we obtain

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

Solution

$I=\int x(\log x)^{2} d x$

Taking $(\log x)^{2}$ as first function and 1 as second function and integrating by parts, we obtain

$ \begin{aligned} I & =(\log x)^{2} \int x d x-\int[\ {\dfrac{d}{d x} (\log x)^{2}\} \int x d x] d x \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\int 2 \log x \cdot \dfrac{1}{x} \cdot \dfrac{x^{2}}{2} d x \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\int x \log x d x \end{aligned} $

Again integrating by parts, we obtain

$ \begin{aligned} I & =\dfrac{x^{2}}{2}(\log x)^{2}-[\log x \int x d x-\int{\dfrac{d}{d x} (\log x) \int x d x} d x] \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-[\dfrac{x^{2}}{2}-\log x-\int \dfrac{1}{x} \cdot \dfrac{x^{2}}{2} d x] \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2} \log x+\dfrac{1}{2} \int x d x \\ & =\dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2} \log x+\dfrac{x^{2}}{4}+C \end{aligned} $

Solution

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

Let $I=I_1+I_2 \qquad…(1)$

Where, $I_1=\int x^{2} \log x d x$ and $I_2=\int \log x d x$

$I_1=\int x^{2} \log x d x$

Taking $\log x$ as first function and $x^{2}$ as second function and integrating by parts, we obtains

$$ \begin{align*} I_1 & =\log x.\int x^{2} d x-\int{\dfrac{d}{d x} (\log x) \int x^{2} d x} d x \\ & =\log x \cdot \dfrac{x^{3}}{3}-\int \dfrac{1}{x} \cdot \dfrac{x^{3}}{3} d x \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{1}{3}\int x^{2} d x \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+C_1 \qquad…(2)\\ I_2 & =\int \log x d x \end{align*} $$

Taking $\log x$ as first function and 1 as second function and integrating by parts, we obtain

$$ \begin{align*} I_2 & =\log x \int 1 \cdot d x-\int{\dfrac{d}{d x} (\log x) \int 1 \cdot d x}dx \\ & =\log x \cdot x-\int \dfrac{1}{x} \cdot x d x \\ & =x \log x-\int 1 d x \\ & =x \log x-x+C_2 \qquad…(3) \end{align*} $$

Using equations (2) and (3) in (1), we obtain

$ \begin{aligned} I & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+C_1+x \log x-x+C_2 \\ & =\dfrac{x^{3}}{3} \log x-\dfrac{x^{3}}{9}+x \log x-x+(C_1+C_2) \\ & =(\dfrac{x^{3}}{3}+x) \log x-\dfrac{x^{3}}{9}-x+C \end{aligned} $

Solution

Let $I=\int e^{x}(\sin x+\cos x) d x$

Let $f(x)=\sin x$

$\Rightarrow f^{’}(x)=\cos x$

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sin x+C$

Solution

Let $I=\int \dfrac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}{\dfrac{x}{(1+x)^{2}}\} d x$

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

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

Let $f(x)=\dfrac{1}{1+x} $

$\Rightarrow f^{\prime}(x)=\dfrac{-1}{(1+x)^{2}}$

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

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

Solution

$e^{x}(\dfrac{1+\sin x}{1+\cos x})$

$=e^{x}(\dfrac{\sin ^{2} \dfrac{x}{2}+\cos ^{2} \dfrac{x}{2}+2 \sin \dfrac{x}{2} \cos \dfrac{x}{2}}{2 \cos ^{2} \dfrac{x}{2}})$

$=\dfrac{e^{x}(\sin \dfrac{x}{2}+\cos \dfrac{x}{2})^{2}}{2 \cos ^{2} \dfrac{x}{2}}$

$=\dfrac{1}{2} e^{x} \cdot \Bigg(\dfrac{\sin \dfrac{x}{2}+\cos \dfrac{x}{2}}{\cos \dfrac{x}{2}}\Bigg)^{2}$

$=\dfrac{1}{2} e^{x}[\tan \dfrac{x}{2}+1]^{2}$

$=\dfrac{1}{2} e^{2}(1+\tan \dfrac{x}{2})^{2}$

$=\dfrac{1}{2} e^{x}[1+\tan ^{2} \dfrac{x}{2}+2 \tan \dfrac{x}{2}]$

$=\dfrac{1}{2} e^{x}[\sec ^{2} \dfrac{x}{2}+2 \tan \dfrac{x}{2}]$

$\dfrac{e^{x}(1+\sin x) d x}{(1+\cos x)}=e^{x}[\dfrac{1}{2} \sec ^{2} \dfrac{x}{2}+\tan \dfrac{x}{2}]\qquad…(1)$

Let $\tan \dfrac{x}{2}=f(x) \Rightarrow f^{\prime}(x)=\dfrac{1}{2} \sec ^{2} \dfrac{x}{2} $

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

From equation (1), we obtain

$\int \dfrac{e^{x}(1+\sin x)}{(1+\cos x)} d x=e^{x} \tan \dfrac{x}{2}+C$

Solution

Let $I=\int e^{x}[\dfrac{1}{x}-\dfrac{1}{x^{2}}] d x$

Also, let $\dfrac{1}{x}=f(x) \Rightarrow f^{\prime}(x)=\dfrac{-1}{x^{2}}$

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)\} d x=e^{x} f(x)+C$

$\therefore I=\dfrac{e^{x}}{x}+C$

Solution

$ \begin{aligned} & \begin{aligned} \int e^{x}{\dfrac{x-3}{(x-1)^{3}}\} d x & =\int e^{x}{\dfrac{x-1-2}{(x-1)^{3}}\} d x \\ & =\int e^{x}{\dfrac{1}{(x-1)^{2}}-\dfrac{2}{(x-1)^{3}}\} d x \end{aligned} \\ & \text{ Let } f(x)=\dfrac{1}{(x-1)^{2}}\ \Rightarrow f^{\prime}(x)=\dfrac{-2}{(x-1)^{3}} \end{aligned} $

It is known that, $\int e^x\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+C$

$\therefore \int e^{x}{\dfrac{(x-3)}{(x-1)^{2}}\} d x=\dfrac{e^{x}}{(x-1)^{2}}+C$

Solution

Let $I=\int e^{2 x} \sin x d x$

Integrating by parts, we obtain

$I=\sin x \int e^{2 x} d x-\int{\dfrac{d}{d x} (\sin x) \int e^{2 x} d x} d x$

$\Rightarrow I=\sin x \cdot \dfrac{e^{2 x}}{2}-\int \cos x \cdot \dfrac{e^{2 x}}{2} d x$

$\Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{1}{2} \int e^{2 x} \cos x d x$

Again integrating by parts, we obtain

$ \begin{aligned} & I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{1}{2}[\cos x \int e^{2 x} d x-\int{\dfrac{d}{d x} (\cos x) \int e^{2 x} d x} d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{1}{2}[\cos x \cdot \dfrac{e^{2 x}}{2}-\int(-\sin x) \dfrac{e^{2 x}}{2} d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{1}{2}[\dfrac{e^{2 x} \cos x}{2}+\dfrac{1}{2} \int e^{2 x} \sin x d x] \\ & \Rightarrow I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4}-\dfrac{1}{4} I \\ & \Rightarrow I+\dfrac{1}{4} I=\dfrac{e^{2 x} \cdot \sin x}{2}-\dfrac{e^{2 x} \cos x}{4} \\ & \Rightarrow \dfrac{5}{4} I=\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4} \\ & \Rightarrow I=\dfrac{4}{5}[\dfrac{e^{2 x} \sin x}{2}-\dfrac{e^{2 x} \cos x}{4}]+C \\ & \Rightarrow I=\dfrac{e^{2 x}}{5}[2 \sin x-\cos x]+C \end{aligned} $

Solution

Let $x=\tan \theta \quad \Rightarrow d x=\sec ^{2} \theta d \theta$

$\therefore \sin ^{-1}(\dfrac{2 x}{1+x^{2}})=\sin ^{-1}(\dfrac{2 \tan \theta}{1+\tan ^{2} \theta})=\sin ^{-1}(\sin 2 \theta)=2 \theta$

$\int \sin^{-1} (\dfrac{2x}{1+x^2})dx=\int 2 \theta \cdot \sec^2 \theta d \theta=2 \int \theta \cdot \sec^2 \theta d \theta$

Integrating by parts, we obtain

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

Solution

Let $I=\int x^{2} e^{x^{3}} d x$

Also, let $x^{3}=t \Rightarrow 3 x^{2} d x=d t$

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

Hence, the correct Answer is A.

Solution

$\int e^{x} \sec x(1+\tan x) d x$

Let $I=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x$

Also, let $\sec x=f(x) \Rightarrow {\sec x \tan x=f^{\prime}(x)}$

It is known that, $\int e^{x}{f(x)+f^{\prime}(x)} d x=e^{x} f(x)+C$

$\therefore I=e^{x} \sec x+C$

Hence, the correct Answer is B.