Question
Evaluate the following integrals:
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
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Answer
$\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
Let $\text{x}\text{e}^\text{x}=\text{t}$
$\Rightarrow\big(1.\text{e}^\text{x}+\text{x}\text{e}^\text{x}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\sec^2\text{t}\text{ dt}$
$=\tan(\text{t})+\text{C}$
$=\tan\big(\text{xe}^\text{x}\big)+\text{C}$
Let $\text{x}\text{e}^\text{x}=\text{t}$
$\Rightarrow\big(1.\text{e}^\text{x}+\text{x}\text{e}^\text{x}\big)=\frac{\text{dt}}{\text{dx}}$
$\Rightarrow(\text{x}+1)\text{e}^\text{x}\text{ dx}=\text{dt}$
Now, $\int\frac{(\text{x}+1)\text{e}^\text{x}}{\cos^2(\text{xe}^\text{x})}\text{ dx}$
$=\int\frac{\text{dt}}{\cos^2\text{t}}$
$=\sec^2\text{t}\text{ dt}$
$=\tan(\text{t})+\text{C}$
$=\tan\big(\text{xe}^\text{x}\big)+\text{C}$
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