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$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}$ equals
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Verified Answer
The correct answer is:
$\tan \left(x e^{x}\right)+\mathrm{C}$
$\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}$
Let $x e^{x}=t$
$\Rightarrow\left(x e^{x}+e^{x}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}$
$\Rightarrow d x=\frac{\mathrm{dt}}{e^{x}(x+1)}$
$\therefore \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}=\int \frac{e^{x}(1+x)}{\cos ^{2} t} \times \frac{d t}{e^{x}(1+x)}=\int \frac{1}{\cos ^{2} t} \mathrm{dt}=\int \sec ^{2} t \mathrm{dt}=$$\tan t+\mathrm{C}=\tan \left(x e^{x}\right)+\mathrm{C}$
Let $x e^{x}=t$
$\Rightarrow\left(x e^{x}+e^{x}\right)=\frac{\mathrm{dt}}{\mathrm{dx}}$
$\Rightarrow d x=\frac{\mathrm{dt}}{e^{x}(x+1)}$
$\therefore \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}=\int \frac{e^{x}(1+x)}{\cos ^{2} t} \times \frac{d t}{e^{x}(1+x)}=\int \frac{1}{\cos ^{2} t} \mathrm{dt}=\int \sec ^{2} t \mathrm{dt}=$$\tan t+\mathrm{C}=\tan \left(x e^{x}\right)+\mathrm{C}$
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