Ex 7.6 Class 12 Maths Ncert Solutions

Let \( \mathrm{I}=x \sin x \)
Now, integrating by parts, we get,
\( \mathrm{I}=x \int \sin x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sin x d x\right\} d x\)
\(=x(-\cos x)-\int 1 \cdot(-\cos x) d x\)
\(=-x \cos x+\sin x+C\)

Let \( \mathrm{I}=x \sin 3 x \)
Now, integrating by parts, we get,
\( \mathrm{I}=x \int \sin 3 x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sin 3 x d x\right\} d x\)
\(=x\left(\frac{-\cos 3 x}{3}\right)-\int 1 \cdot\left(\frac{-\cos 3 x}{3}\right) d x\)
\(=\frac{-x \cos 3 x}{3}+\frac{1}{9} \sin 3 x+C\)

Let \( \mathrm{I}=x^{2} e^{x} \)
Now, integrating by parts, we get,
\( \mathrm{I}=x^{2} \int e^{x} d x-\int\left\{\left(\frac{d}{d x} x^{2}\right) \int e^{x} d x\right\} d x\)
\(=x^{2} e^{x}-\int 2 x \cdot e^{x} d x\)
\(=x^{2} e^{x}-2 \int x \cdot e^{x} d x\)
Again integrating by parts, we get,
\(=x^{2} e^{x}-2\left[x \int e^{x} d x-\int\left\{\left(\frac{d}{d x} x^{2}\right) \int e^{x} d x\right\} d x\right]\)
\(=x^{2} e^{x}-2\left[x e^{x}-\int e^{x} d x\right]\)
\(=x^{2} e^{x}-2\left[x e^{x}-e^{x}\right]\)
\(=x^{2} e^{x}-2 x e^{x}+e^{x}+C\)
\(=e^{x}\left(x^{2}-2 x+2\right)+C\)

Let \( \mathrm{I}=x \log x \)
Now, integrating by parts, we get,
Taking, Logarithmic function as first function and algebraic function as second function,
\( \mathrm{I}=\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x\)
\(=\log x\left(\frac{x^{2}}{2}\right)-\int \frac{1}{2} \cdot \frac{x^{2}}{2} d x\)
\(=\frac{x^{2} \log x}{2}-\int \frac{x}{2} d x\)
\(=\frac{x^{2} \log x}{2}-\frac{x^{2}}{4}+C\)

Let \( \mathrm{I}=x\log 2x \)
Now, integrating by parts, we get,
\( \mathrm{I}=\log 2 x \int x d x-\int\left\{\left(\frac{d}{d x} 2 \log x\right) \int x d x\right\} d x\)
\(=\log 2 x\left(\frac{x^{2}}{2}\right)-\int \frac{2}{2 x} \cdot \frac{x^{2}}{2} d x\)
\(=\frac{x^{2} \log 2 x}{2}-\int \frac{x}{2} d x\)
\(=\frac{x^{2} \log 2 x}{2}-\frac{x^{2}}{4}+C\)

Let \( \mathrm{I}=x^{2} \log x \)
Now, integrating by parts, we get,
\( \mathrm{I}=\log x \int x^{2} d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x^{2} d x\right\} d x\)
\(=\log x\left(\frac{x^{3}}{3}\right)-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x\)
\(=\frac{x^{3} \log x}{2}-\int \frac{x^{2}}{2} d x\)
\(=\frac{x^{3} \log x}{2}-\frac{x^{3}}{4}+C\)

Let \( \mathrm{I}=x \sin ^{-1} x \)
Now, integrating by parts, we get,
\( \mathrm{I}=\sin ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \sin ^{-1} x\right) \int x d x\right\} d x\)
\(=\sin ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x\)
\(=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int \frac{-x^{2}}{\sqrt{1-x^{2}}} d x\)
\(=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\frac{1-x^{2}}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x\)
\(=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2} \int\left\{\sqrt{1-x^{2}}-\frac{1}{\sqrt{1-x^{2}}}\right\} d x\)
\(=\frac{x^{2} \sin ^{-1} x}{2}+\frac{1}{2}\left\{\int \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}}\right\} d x\)
\(=\frac{x^{2} \sin ^{-1} x}{2}+\frac{x}{4} \sqrt{1-x^{2}}+\frac{1}{4} \sin ^{-1} x-\frac{1}{2} \sin ^{-1} x+C\)
\(=\frac{1}{4}\left(2 x^{2}-1\right) \sin ^{-1} x+\frac{x}{4} \sqrt{1-x^{2}}+\mathrm{C}\)

Let \( \mathrm{I}=x \tan ^{-1} x \)
Now, integrating by parts, we get,
\( \mathrm{I}=\tan ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int x d x\right\} d x\)
\(=\tan ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x\)
\(=\frac{x^{2} \tan ^{-1} x}{2}+\frac{1}{2} \int \frac{x^{2}}{\sqrt{1-x^{2}}} d x\)
\(=\frac{x^{2} \tan ^{-1} x}{2}+\frac{1}{2} \int\left(\frac{x^{2}+1}{1+x^{2}}-\frac{1}{1+x^{2}}\right) d x\)
\(=\frac{x^{2} \tan ^{-1} x}{2}+\frac{1}{2} \int\left\{1-\frac{1}{1+x^{2}}\right\} d x\)
\(=\frac{x^{2} \tan ^{-1} x}{2}+\frac{1}{2}\left(x-\tan ^{-1} x\right)+C\)
\(=\frac{x^{2} \tan ^{-1} x}{2}-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+C\)

Let \( \mathrm{I}=x \cos ^{-1} x \)
Now, integrating by parts, we get,
\( \mathrm{I}=\cos ^{-1} x \int x d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int x d x\right\} d x\)
\(=\cos ^{-1} x \cdot \frac{x^{2}}{2}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot \frac{x^{2}}{2} d x\)
\(=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \frac{-1-x^{2}-1}{\sqrt{1-x^{2}}} d x\)
\(=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int\left\{\sqrt{1-x^{2}}+\left(\frac{-1}{\sqrt{1-x^{2}}}\right)\right\} d x\)
\(=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \int \sqrt{1-x^{2}}+\left(\frac{-1}{\sqrt{1-x^{2}}}\right) d x\)
\(=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2} \mathrm{I}_{1}-\frac{1}{2} \cos ^{-1} x \ldots(1)\)
Now, \( \mathrm{I}_{1}=\int \sqrt{1-x^{2}} d x \)
\( \mathrm{I}_{1}=x \sqrt{1-x^{2}}-\int \frac{d}{d x} \sqrt{1-x^{2}} \int x d x\)
\( \mathrm{I}_{1}=x \sqrt{1-x^{2}}-\int \frac{-2 x}{2 \sqrt{1-x^{2}}} x \cdot d x\)
\(=x \sqrt{1-x^{2}}-\int \frac{-x^{2}}{\sqrt{1-x^{2}}} x \cdot d x\)
\(=x \sqrt{1-x^{2}}-\int \frac{1+x^{2}-1}{\sqrt{1-x^{2}}} x \cdot d x\)
\(=x \sqrt{1-x^{2}}-\left\{\int \sqrt{1-x^{2}} d x+\int \frac{d x}{\sqrt{1-x^{2}}}\right\}\)
\( \mathrm{I}_{1}=x \sqrt{1-x^{2}}-\left\{I_{1}+\cos ^{-1} x\right\}\)
\( \mathrm{I}_{2}=\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x\)
Now, substituting in (1), we get,
\( \mathrm{I}=\frac{x^{2} \cos ^{-1} x}{2}-\frac{1}{2}\left(\frac{x}{2} \sqrt{1-x^{2}}-\frac{1}{2} \cos ^{-1} x\right)-\frac{1}{2} \cos ^{-1} x\)
\(=\frac{\left(2 x^{2}-1\right)}{4} \cos ^{-1} x-\frac{x}{4} \sqrt{1-x^{2}}+C\)

Let \( \mathrm{I}=\left(\sin ^{-1} x\right)^{2} \)
Now, integrating by parts, we get,
\(\mathrm{I}=\left(\sin ^{-1} x\right) \int x d x-\int\left\{\left(\frac{d}{d x}\left(\sin ^{-1} x\right)^{2}\right) \cdot \int x d x\right\} d x\)
\(=\left(\sin ^{-1} x\right)^{2} \cdot x-\int \frac{2 \sin ^{-1} x}{\sqrt{1-x^{2}}} \cdot x d x\)
\(=x\left(\sin ^{-1} x\right)^{2}+\int \sin ^{-1} x\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right) d x\)
\(=x\left(\sin ^{-1} x\right)^{2}+\left[\sin ^{-1} x \int\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right) d x-\int\left\{\left(\frac{d}{d x}\left(\sin ^{-1} x\right)^{2}\right) \right.\right.\)\(\left.\left.\int\left(\frac{-2 x}{\sqrt{1-x^{2}}}\right) d x\right\} d x\right]\)
\(=x\left(\sin ^{-1} x\right)^{2}+\left[\sin ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{1}{\sqrt{1-x^{2}}} \cdot 2\left(\sqrt{1-x^{2}}\right) d x\right]\)
\(=x\left(\sin ^{-1} x\right)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-\int 2 d x\)
\(=x\left(\sin ^{-1} x\right)^{2}+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+C\)

Let \( \mathrm{I}=\frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} \)
\(\mathrm{I}=-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cdot \cos ^{-1} x d x\)
Now, integrating by parts, we get,
\(\mathrm{I}=-\frac{1}{2} \int \cos ^{-1} x \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\)
\(=-\frac{1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int \frac{-1}{\sqrt{1-x^{2}}} \cdot 2 \sqrt{1-x^{2}} d x\right]\)
\(=-\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+\int 2 d x\right]\)
\(=\frac{1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+\mathrm{C}\)
\(=-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+\mathrm{C}\)

Let \( \mathrm{I}=x \sec ^{2} x \)
Now, integrating by parts, we get,
\(\mathrm{I}=x \int \sec ^{2} x d x-\int\left\{\left(\frac{d}{d x} x\right) \int \sec ^{2} x d x\right\} d x\)
\(=x \tan x-\int 1 \tan x d x\)
\(=x \tan x+\log |\cos x|+C\)

Let \( \mathrm{I}=\tan ^{-1} x \)
So, now integrating by parts, we get,
\(\mathrm{I}=\tan ^{-1} x \int 1 \cdot d x-\int\left\{\left(\frac{d}{d x} \tan ^{-1} x\right) \int 1 \cdot d x\right\} d x\)
\(=\tan ^{-1} x \cdot x-\int \frac{1}{1+x^{2}} x \cdot d x\)
\(=x \tan ^{-1} x-\frac{1}{2} \int \frac{2 x}{1+x^{2}} d x\)
\(=x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^{2}\right|+C\)
\(=x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+C\)

Let \( \mathrm{I}=x(\log x)^{2} \)
Integrating by parts, we get,
\(\mathrm{I}=\frac{x^{2}}{2}(\log x)^{2}-\left[\log x \int x d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x\right]\)
\(=\frac{x^{2}}{2}(\log x)^{2}-\left[\frac{x^{2}}{2}-\log x-\int \frac{1}{x} \cdot \frac{x^{3}}{2} d x\right]\)
\(=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{1}{2} \int x \cdot d x\)
\(=\frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+C\)

Let \( \mathrm{I}=\int\left(x^{2}+1\right) \log x d x=\int x^{2} \log x d x+\int \log x d x \)
Now, Let \( \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2} \ldots \) (1)
Where, \( \mathrm{I}_{1}=\int x^{2} \log x d x \) and \( \mathrm{I}_{2}=\int \log x d x \)
So, now
\(\mathrm{I}_{1}=\int x^{2} \log x d x\)
Integrating by parts, we get,
\(\mathrm{I}_{1}=\log x-\int x^{2} d x-\iint\left\{\left(\frac{d}{d x} \log x\right) \int x d x\right\} d x\)
\(=\log x \cdot \frac{x^{3}}{3}-\int \frac{1}{x} \cdot \frac{x^{3}}{3} d x\)
\(=\frac{x^{3}}{3} \log x-\frac{1}{3}\left(\int x^{2} \cdot d x\right)\)
\(=\frac{x^{3}}{3} \log x \frac{x^{3}}{9}+C_{1} \ldots(2)\)
Now, \( \mathrm{I}_{2}=\int \log x d x \)
Integrating by parts, we get,
\(\mathrm{I}_{2}=\log x \int 1 \cdot d x-\int\left\{\left(\frac{d}{d x} \log x\right) \int 1 \cdot d x\right\}\)
\(=\log x \cdot x-\int \frac{1}{x} \cdot x d x\)
\(=x \log x-\int 1 . d x\)
\(=x \log x+C_{2} \ldots(3\)
Now putting the value of \( \mathrm{I}_{1} \) and \( \mathrm{I}_{2} \), in (1), we get,
\(\mathrm{I}=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+C_{1}+x \log x-x+C_{2}\)
\(=\frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+x \log x-x+\left(C_{1}+C_{2}\right)\)
\(=\left(\frac{x^{3}}{3}+x\right) \log x-\frac{x^{3}}{9}-x+C\)

\(\mathrm{I}=\int e^{x}(\sin x+\cos x) d x\)
Now,
Let \( \sin x=f(x) \)
\(\Rightarrow f^{\prime}(x)=\cos x \)
We know that,
\(\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C\)
Thus,
\(\int e^{x}(\sin x+\cos x) d x=e^{x} \sin x+C\)

\(\mathrm{I}=\int \frac{x e^{x}}{(1+x)^{2}} d x=\int e^{x}\left\{\frac{x}{(1+x)^{2}}\right\} d x\)
\(=\int e^{x}\left\{\frac{1+x-1}{(1+x)^{2}}\right\} d x\)
\(\int e^{x}\left\{\frac{1}{1+x}-\frac{1}{(1+x)^{2}}\right\} d x\)
Now,
Let \( \frac{1}{1+x}=f(x) \)
\(\Rightarrow f^{\prime}(x)=\frac{1}{(1+x)^{2}} \)
We know that,
\(\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C\)
Thus,
\(\int \frac{x e^{x}}{(1+x)^{2}} d x=\frac{e^{x}}{1+x}+\mathrm{C}\)

\(e^{x}\left(\frac{1+\sin x}{1+\cos x}\right)\)
\(=e^{x}\left(\frac{\sin ^{2} \frac{x}{2} \cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}}\right)\)
\(=\frac{e^{x}\left(\sin \frac{x}{2} \cos \frac{x}{2}\right)^{2}}{2 \cos ^{2} \frac{x}{2}}\)
\(=\frac{1}{2} e^{x} \cdot\left(\frac{\sin \frac{x}{2} \cos \frac{x}{2}}{\cos \frac{x}{2}}\right)^{2}\)
\(=\frac{1}{2} e^{x}\left(\tan \frac{x}{2}+1\right)^{2}\)
\(=\frac{1}{2} e^{x}\left[1+\tan ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right]\)
\(=\frac{1}{2} e^{x}\left[\sec ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right]\)
\( \Rightarrow e^{x}\left[\frac{1+\sin x}{1+\cos x}\right]=\frac{1}{2} e^{x}\left[\sec ^{2} \frac{x}{2}+2 \tan \frac{x}{2}\right] \)
Now,
Let \( \tan \frac{x}{2}=f(x) \)
\(\Rightarrow f^{\prime}(x)=\frac{1}{2} \sec ^{2} \frac{x}{2} \)
We know that,
\(\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C\)
Thus,
\(\int e^{x}\left[\frac{1+\sin x}{1+\cos x}\right] d x=e^{x} \tan \frac{x}{2}+C\)

Let \( \mathrm{I}=\int e^{x}\left[\frac{1}{x}-\frac{1}{x^{2}}\right] d x \)
Now, let
\(\frac{1}{x}=f(x) \)
\(\Rightarrow f^{\prime}(x)=\frac{1}{x^{2}}\)
We know that,
\(\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C\)
Thus,
\(\mathrm{I}=\frac{e^{x}}{x}+\mathrm{C}\)

\(\int e^{x}\left[\frac{(x-3)}{(x-1)^{3}}\right] d x=\int e^{x}\left\{\frac{x-1-2}{(x-1)^{3}}\right\} d x\)
\(=\int e^{x}\left\{\frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\right\}\)
Now, let \( \mathrm{f}(x)=\frac{1}{(x-1)^{2}} \)
\(\Rightarrow f^{\prime}(x)=\frac{2}{(x-1)^{3}}\)
We know that,
\(\int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C\)
Thus,
\(\int e^{x}\left\{\frac{(x-3)}{(x-1)^{3}}\right\} d x=\frac{e^{x}}{(x-1)^{2}}+C\)

Let \( \mathrm{I}=e^{2 x} \sin x \)
Integrating by parts, we get,
\(\mathrm{I}=\sin x \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x\)
\(=\sin x \cdot \frac{e^{2 x}}{2}-\int \cos x \cdot \frac{e^{2 x}}{2} d x\)
Again, integrating by parts, we get,
\(\mathrm{I}=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos \int e^{2 x} d x-\int\left\{\left(\frac{d}{d x} \sin x\right) \int e^{2 x} d x\right\} d x\right]\)
\(=\frac{e^{2 x} \sin x}{2}-\frac{1}{2}\left[\cos x \cdot \frac{e^{2 x}}{2}-\int(-\sin x) \cdot \frac{e^{2 x}}{2} d x\right]\)
\(=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{2}-\frac{1}{4} \mathrm{I}\)
\(\Rightarrow \mathrm{I}+\frac{1}{4} \mathrm{I}=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{2}\)
\(\Rightarrow \frac{5}{4} \mathrm{I}=\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{2}\)
\(\Rightarrow \frac{4}{5}\left[\frac{e^{2 x} \sin x}{2}-\frac{e^{2 x} \cos x}{2}\right]\)

Let \( x=\tan \theta \)
\(\Rightarrow \mathrm{d}x=\sec ^{2} \theta d \theta\)
Thus,
\(\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan \theta}\right)=\sin ^{-1}(\sin 2 \theta)=2 \theta\)
\(\Rightarrow \int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x=\int 2 \theta \mathrm{d} \theta=2 \int \theta \sec ^{2} \theta \mathrm{d} \theta\)
Integrating by parts, we get
\(2\left[\theta \cdot \int \theta \sec ^{2} \theta \mathrm{d} \theta-\int\left\{\left(\frac{d}{d \theta}\right) \int \sec ^{2} \theta \mathrm{d} \theta\right\} \mathrm{d} \theta\right]\)
\(=2\left[\theta \cdot \tan \theta-\int \tan \theta d \theta\right]\)
\(=2[\theta \tan \theta+\log |\cos \theta|+\mathrm{C}]\)
\(=2\left[x \tan ^{-1} x+\log \left|\frac{1}{\sqrt{1+x^{2}}}\right|+C\right]\)
\(=2 x \tan ^{-1} x+2 \log \left(1+x^{2}\right)^{\frac{1}{2}}+C\)
\(=2 x \tan ^{-1} x+2\left[-\frac{1}{2} \log (1+x)^{2}\right]+C\)
\(=2 x \tan ^{-1} x+\log \left(1+x^{2}\right)+C\)

Let \( \mathrm{I}=\int x^{2} e^{x^{3}} d x \)
Also, let \( x^{3}=\mathrm{t} \)
\(\Rightarrow 3 x^{2} \mathrm{d}x=\mathrm{dt}\)
Thus,
\(\Rightarrow \mathrm{I}=\frac{1}{3} \int e^{t} d t\)
\(\Rightarrow \frac{1}{3}\left(e^{t}\right)+C\)
\(\Rightarrow \frac{1}{3}\left(e^{x^{3}}\right)+C\)

\(\int e^{x} \sec x(1+\tan x) d x\)
Let \( \mathrm{I}=\int e^{x} \sec x(1+\tan x) d x=\int e^{x}(\sec x+\sec x \tan x) d x \)
Also, let \( \sec x=\mathrm{f}(x) \)
\(\Rightarrow \sec x \tan x=\mathrm{f}^{\prime}(x)\)
We know that, \( \int e^{x}\left\{f(x)+f^{\prime}(x)\right\} d x=e^{x} f(x)+C \)
Thus, \( \mathrm{I}=e^{x} \sec x+\mathrm{C} \)