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Question: \[\int_{}^{}\frac{1 + x + \sqrt{x + x^{2}}}{\sqrt{x} + \sqrt{1 + x}}dx =\]...

∫1+x+x+x2x+1+xdx=\int_{}^{}\frac{1 + x + \sqrt{x + x^{2}}}{\sqrt{x} + \sqrt{1 + x}}dx =

A

121+x+c\frac{1}{2}\sqrt{1 + x} + c

B

23(1+x)3/2+c\frac{2}{3}(1 + x)^{3/2} + c

C

1+x+c\sqrt{1 + x} + c

D

2(1+x)3/2+c2(1 + x)^{3/2} + c

Answer

23(1+x)3/2+c\frac{2}{3}(1 + x)^{3/2} + c

Explanation

Solution

∫1+x+x+x2x+1+xdx\int_{}^{}\frac{1 + x + \sqrt{x + x^{2}}}{\sqrt{x} + \sqrt{1 + x}}dx =∫1+x[1+x+x](x+1+x)dx= \int_{}^{}\frac{\sqrt{1 + x}\left\lbrack \sqrt{1 + x} + \sqrt{x} \right\rbrack}{\left( \sqrt{x} + \sqrt{1 + x} \right)}dx

=∫1+xdx=23(1+x)3/2+c\mathbf{=}\int_{}^{}\sqrt{\mathbf{1 + x}}\mathbf{dx =}\frac{\mathbf{2}}{\mathbf{3}}\mathbf{(1 + x}\mathbf{)}^{\mathbf{3/2}}\mathbf{+ c}