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$\int \frac{e^x(1+x)}{\cos ^2\left(e^x \cdot x\right)} d x$ equal
(a) $-\cot \left(e \cdot x^x\right)+C$
(b) $\tan \left(x e^x\right)+C$
(c) $\tan \left(e^x\right)+C$
(d) $\cot \mathrm{e}^{\mathrm{x}}+\mathrm{C}$
$\int \frac{e^x(1+x)}{\cos ^2\left(e^x \cdot x\right)} d x$ equal
(a) $-\cot \left(e \cdot x^x\right)+C$
(b) $\tan \left(x e^x\right)+C$
(c) $\tan \left(e^x\right)+C$
(d) $\cot \mathrm{e}^{\mathrm{x}}+\mathrm{C}$
Solution:
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Verified Answer
(b) $\int \frac{e^x(1+x)}{\cos ^2\left(e^x \cdot x\right)} d x=\int \frac{d t}{\cos ^2 t}$
Let $x e^x=t \Rightarrow\left(e^x \cdot 1+e^x x\right) d x=d t$ $=\int \sec ^2 t d t=\tan \mathrm{t}+\mathrm{C}=\tan \left(x e^{\mathrm{x}}\right)+\mathrm{C}$
Let $x e^x=t \Rightarrow\left(e^x \cdot 1+e^x x\right) d x=d t$ $=\int \sec ^2 t d t=\tan \mathrm{t}+\mathrm{C}=\tan \left(x e^{\mathrm{x}}\right)+\mathrm{C}$
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