Question

$\int \frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx=$

• A. $\frac{1}{2}\sqrt{1+x}+c$
• B. $\frac{1}{2}(1+x{)}^{3/2}+c$
• C. $\sqrt{1+x}+c$
• D. $\frac{2}{3}(1+x{)}^{3/2}+c$

Answer 1

The correct option is D $\frac{2}{3}(1+x{)}^{3/2}+c$

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

$=\int \frac{\sqrt{1+x}(\sqrt{1+x}+\sqrt{x})}{(\sqrt{1+x}+\sqrt{x})}dx=\int \sqrt{1+x}dx$

Let $1+x=z^2\Rightarrow dx=2zdz$

$\therefore \int \sqrt{1+x}dx=\int z.2zdz=2\int z^{2}dz=\frac{2}{3}{z}^3+c$

$=\frac{2}{3}(1+x{)}^{\frac{3}{2}}+C$, $c=$ constant of integration.