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

Solution

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

Answer

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\{(\frac{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\{(\frac{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\{(\frac{d}{d x} \log x) \int x d x\} d x \\ & =\log x \cdot \frac{x^{2}}{2}-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x \\ & =\frac{x^{2} \log x}{2}-\frac{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\{(\frac{d}{d x} 2 \log x) \int x d x\} d x \\ & =\log 2 x \cdot \frac{x^{2}}{2}-\int \frac{2}{2 x} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \log 2 x}{2}-\int \frac{x}{2} d x \\ & =\frac{x^{2} \log 2 x}{2}-\frac{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\{(\frac{d}{d x} \log x) \int x^{2} d x\} d x \\ & =\log x(\frac{x^{3}}{3})-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x \\ & =\frac{x^{3} \log x}{3}-\int \frac{x^{2}}{3} d x \\ & =\frac{x^{3} \log x}{3}-\frac{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\{(\frac{d}{d x} \sin ^{-1} x) \int x d x\} d x \\ & =\sin ^{-1} x(\frac{x^{2}}{2})-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x \\ & =\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int [\frac{1-x^2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}} ]d x \\ & =\frac{x^2 \sin ^{-1} x}{2}+\frac{1}{2} \int [\sqrt{1-x^2}-\frac{1}{\sqrt{1-x^2}} ] d x\\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\{\int \sqrt{1-x^{2}} d x-\int \frac{1}{\sqrt{1-x^{2}}} d x\} \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\{\frac{x}{2} \sqrt{1-x^{2}}+\frac{1}{2} \sin ^{-1} x-\sin ^{-1} x\}+C \\ & =\frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C \\ & =\frac{1}{4}(2 x^{2}-1) \sin ^{-1} x+\frac{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\{(\frac{d}{d x} \tan ^{-1} x) \int x d x\} d x \\ & =\tan ^{-1} x(\frac{x^{2}}{2})-\int \frac{1}{1+x^{2}} \cdot \frac{x^{2}}{2} d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int \frac{x^{2}}{1+x^{2}} d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int(\frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}}) d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2} \int(1-\frac{1}{1+x^{2}}) d x \\ & =\frac{x^{2} \tan ^{-1} x}{2}-\frac{1}{2}(x-\tan ^{-1} x)+C \\ & =\frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{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

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

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

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

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

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

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

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

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

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

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

Substituting in (1), we obtain

$ \begin{aligned} I & =\frac{x \cos ^{-1} x}{2}-\frac{1}{2}(\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x)-\frac{1}{2} \cos ^{-1} x \\ & =\frac{(2 x^{2}-1)}{4} \cos ^{-1} x-\frac{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) \int 1 d x-\int\{\frac{d}{d x}(\sin ^{-1} x)^{2} \cdot \int 1 \cdot d x\} d x \\ & =(\sin ^{-1} x)^{2} \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x \\ & =x(\sin ^{-1} x)^{2}+\int \sin ^{-1} x \cdot(\frac{-2 x}{\sqrt{1-x^{2}}}) d x \\ & =x(\sin ^{-1} x)^{2}+[\sin ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\{(\frac{d}{d x} \sin ^{-1} x) \int \frac{-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 \frac{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 $I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$ $I=\frac{-1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x$

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

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

Again integrating by parts, we obtain

$ \begin{aligned} I & =\frac{x^{2}}{2}(\log x)^{2}-[\log x \int x d x-\int\{(\frac{d}{d x} \log x) \int x d x\} d x] \\ & =\frac{x^{2}}{2}(\log x)^{2}-[\frac{x^{2}}{2}-\log x-\int \frac{1}{x} \cdot \frac{x^{2}}{2} d x] \\ & =\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{1}{2} \int x d x \\ & =\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{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 $I=I_1+I_2 \ldots$ (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 obtain

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

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

$$ \begin{align*} I_2 & =\log x \int 1 \cdot d x-\int\{(\frac{d}{d x} \log x) \int 1 \cdot d x\} \\ & =\log x \cdot x-\int \frac{1}{x} \cdot x d x \\ & =x \log x-\int 1 d x \\ & =x \log x-x+C_2 \tag{3} \end{align*} $$

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

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

Solution

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

Let $f(x)=\sin x$

$e^x\left(f(x)+f^{\prime}(x)\right)^{\prime}(x)=\cos x$

$e^x\left(f(x)+f^{\prime}(x)\right) {I=\int e^{x}}\{f(x)+f^{\prime}(x)\} d x$

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

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

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

Let $f(x)=\frac{1}{1+x} e^x\left(f(x)+f^{\prime}(x)\right) f^{\prime}(x)=\frac{-1}{(1+x)^{2}}$

$\Rightarrow \int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\{f(x)+f^{\prime}(x)\} d x$

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

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

Solution

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

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

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

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

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

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

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

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

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

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

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

From equation (1), we obtain

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

Solution

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

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

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

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

Solution

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

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

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

Solution

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

Integrating by parts, we obtain

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

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

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

Again integrating by parts, we obtain

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

Solution

Let $x=\tan \theta \quad \square d x=\sec ^{2} \theta d \theta$ $\therefore \sin ^{-1}(\frac{2 x}{1+x^{2}})=\sin ^{-1}(\frac{2 \tan \theta}{1+\tan ^{2} \theta})=\sin ^{-1}(\sin 2 \theta)=2 \theta$

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

Integrating by parts, we obtain

$ \begin{aligned} & 2[\theta \cdot \int \sec ^{2} \theta d \theta-\int\{(\frac{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 |\frac{1}{\sqrt{1+x^{2}}}|]+C \\ & =2 x \tan ^{-1} x+2 \log (1+x^{2})^{-\frac{1}{2}}+C \\ & =2 x \tan ^{-1} x+2[-\frac{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 $x^{3}=t3 x^{2} d x=d t$

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