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$\int \frac{d x}{1+\sqrt{x}}=$
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Verified Answer
The correct answer is:
$2 \sqrt{x}-2 \log |1+\sqrt{x}|+c$
(C)
Let $\quad \mathrm{I}=\int \frac{\mathrm{dx}}{1+\sqrt{\mathrm{x}}}$
Put $\quad \sqrt{\mathrm{x}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=2 \mathrm{t} \mathrm{dt}$
$\therefore \mathrm{I}=\int \frac{2 \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}$
$\quad=2 \int \frac{(\mathrm{t}+1)-1}{1+\mathrm{t}} \mathrm{dt}=2 \int \mathrm{dt}-2 \int \frac{\mathrm{dt}}{1+\mathrm{t}}$
$\quad=2 \mathrm{t}-2 \log |1+\mathrm{t}|=2 \sqrt{\mathrm{x}}-2 \log |1+\sqrt{\mathrm{x}}|+\mathrm{c}$
Let $\quad \mathrm{I}=\int \frac{\mathrm{dx}}{1+\sqrt{\mathrm{x}}}$
Put $\quad \sqrt{\mathrm{x}}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\mathrm{x}}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=2 \mathrm{t} \mathrm{dt}$
$\therefore \mathrm{I}=\int \frac{2 \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}$
$\quad=2 \int \frac{(\mathrm{t}+1)-1}{1+\mathrm{t}} \mathrm{dt}=2 \int \mathrm{dt}-2 \int \frac{\mathrm{dt}}{1+\mathrm{t}}$
$\quad=2 \mathrm{t}-2 \log |1+\mathrm{t}|=2 \sqrt{\mathrm{x}}-2 \log |1+\sqrt{\mathrm{x}}|+\mathrm{c}$
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