A) \[\frac{2}{3}{{(1+x)}^{2/3}}-\frac{2}{3}{{x}^{2/3}}+c\]
B) \[\frac{3}{2}{{(1+x)}^{2/3}}+\frac{3}{2}{{x}^{2/3}}+c\]
C) \[\frac{3}{2}{{(1+x)}^{3/2}}+\frac{3}{2}{{x}^{3/2}}+c\]
D) \[\frac{2}{3}{{(1+x)}^{3/2}}-\frac{2}{3}{{x}^{3/2}}+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{dx}{\sqrt{1+x}+\sqrt{x}}=\int_{{}}^{{}}{\left[ \frac{(x+1)-x}{\sqrt{1+x}+\sqrt{x}} \right]\,dx}}\] \[\int_{{}}^{{}}{(\sqrt{x+1}-\sqrt{x})\,dx}=\frac{{{(x+1)}^{3/2}}}{3/2}-\frac{{{x}^{3/2}}}{3/2}+c\] \[=\frac{2}{3}[{{(x+1)}^{3/2}}-{{x}^{3/2}}]+c=\frac{2}{3}{{(x+1)}^{3/2}}-\frac{2}{3}{{x}^{3/2}}+c.\]
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