Integral Calculus Question 313
Question: $ \int_{{}}^{{}}{\frac{dx}{\sqrt{1+x}+\sqrt{x}}=} $
Options:
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 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{dx}{\sqrt{1+x}+\sqrt{x}}=\int_{{}}^{{}}{[ \frac{(x+1)-x}{\sqrt{1+x}+\sqrt{x}} ],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. $
BETA
