∫ e^ x (1+x) ^ 2 (e^ x x) dx equals

\int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} \mathrm{dx}

equals

Options

A-\cot $\left(e x^{x}\right)+\mathrm{C}$
B\tan $\left(x e^{x}\right)+\mathrm{C}$
C\tan $\left(e^{x}\right)+\mathrm{C}$
D\cot $\left(e^{x}\right)+\mathrm{C}$

Correct answer

B. \tan $\left(x e^{x}\right)+\mathrm{C}$

Step-by-step solution

\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}

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