Question

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx=A(1+x{)}^{\frac{3}{2}}+c$ then A =

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $1$
• D. $-\frac{3}{2}$

Answer 1

The correct option is B $\frac{2}{3}$

$\int \frac{(\sqrt{1+x}{)}^2+\sqrt{x}\sqrt{1+x}dx}{\sqrt{x}+\sqrt{1+x}}=\int \sqrt{1+x}\frac{(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})}dx=\frac{(1+x{)}^{\frac{3}{2}}}{\frac{3}{2}}+C\therefore A=\frac{2}{3}$