Question

The value of the definite integral ${\int }_0^1\frac{1+x+\sqrt{x+x^2}dx}{\sqrt{x}+\sqrt{1+x}}dx$ is?

• A. $\frac{1}{3}(2^{1/2}-1)$
• B. $\frac{2}{3}(2^{1/2}-1)$
• C. $\frac{2}{3}(2^{3/2}-1)$
• D. $\frac{1}{3}(2^{3/2}-1)$

Answer 1

Answer: (C)

$\frac{2}{3}(2^{3/2}-1)$

$I={\int }_0^1\frac{1+x+\sqrt{x+x^2}}{\sqrt{x}+\sqrt{1+x}}dx$

$\therefore I={\int }_0^1\frac{\sqrt{1+x}(\sqrt{1+x}+\sqrt{x})}{\sqrt{x}+\sqrt{1+x}}dx$

$\therefore I={\int }_0^1\sqrt{1+x}dx$

Put $x={\tan }^{2}t⟹dx=2\tan t{\sec }^{2}tdt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\sqrt{1+{\tan }^{2}t}2\tan t{\sec }^{2}tdt$

$\therefore I={\int }_0^{\frac{\pi }{4}}\sqrt{{\sec }^{2}t}2\tan t{\sec }^{2}tdt$

$\therefore I={\int }_0^{\frac{\pi }{4}}2\tan t{\sec }^{3}tdt$

Put $\sec t=y⟹\sec t\tan tdt=dy$

$\therefore I={\int }_1^{\sqrt{2}}2y^{2}dy$

$\therefore I=\frac{2}{3}{y}^3{|}_1^{\sqrt{2}}$

$\therefore I=\frac{4\sqrt{2}-2}{3}$

$\therefore I=\frac{2}{3}(2^{\frac{3}{2}}-1)$