$\int \frac{\sqrt{x} \, dx}{1 + x}$ equals
A$\log \left(\frac{1 + \sqrt{x}}{\sqrt{x}}\right) + c$
B$\log \left(\frac{\sqrt{x}}{1 + x}\right) + c$
C$2\sqrt{x} - 2\tan^{-1}\sqrt{x} + c$
D$2\sqrt{x} - \tan^{-1}\sqrt{x} + c$
Answer & Solution
Correct answer: C. $2\sqrt{x} - 2\tan^{-1}\sqrt{x} + c$
Let $\sqrt{x}=t$. Then $x=t^2$ and $dx=2t\,dt$.
So the integral becomes
$$\int \frac{\sqrt{x}}{1+x}\,dx=\int \frac{t}{1+t^2}\cdot 2t\,dt$$
$$=\int \frac{2t^2}{1+t^2}\,dt$$
$$=\int 2\left(1-\frac{1}{1+t^2}\right)dt$$
$$=2t-2\tan^{-1}t+c$$
Substituting back $t=\sqrt{x}$,
$$2\sqrt{x}-2\tan^{-1}\sqrt{x}+c$$
This matches option $\mathrm{(C)}$.
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