Q71 / 125 · BITSAT 2021 · Mathematics

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

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

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}$

Hence, the correct answer is option: 'B'.

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