Question

1∫ 1/√1+x-√x dx ,Evaluate it.0

Answer 1

Answer:

$\int{\frac{1}{\sqrt{1+x}-\sqrt{x}}}\:dx=\frac{2}{3}}[(1+x)^{\frac{3}{2}}+x^{\frac{3}{2}}]+c$

Step-by-step explanation:

$\int{\frac{1}{\sqrt{1+x}-\sqrt{x}}}\:dx$

$=\int{\frac{1}{\sqrt{1+x}-\sqrt{x}}}\times\frac{\sqrt{1+x}+\sqrt{x}}{\sqrt{1+x}+\sqrt{x}}\:dx$

using

$\boxed{(a+b)(a-b)=a^2-b^2}$

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

$=\int{\frac{\sqrt{1+x}+\sqrt{x}}{1+x-x}\:dx$

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

$=\int{[(1+x)^{\frac{1}{2}}+x^{\frac{1}{2}}]\:dx$

$=\frac{(1+x)^{\frac{3}{2}}}{\frac{3}{2}}}+\frac{x^{\frac{3}{2}}}{\frac{3}{2}}}+c$

$=\frac{2}{3}}(1+x)^{\frac{3}{2}}+\frac{2}{3}}x^{\frac{3}{2}}+c$

$=\frac{2}{3}}[(1+x)^{\frac{3}{2}}+x^{\frac{3}{2}}]+c$

$\implies\:\boxed{\int{\frac{1}{\sqrt{1+x}-\sqrt{x}}\:dx=\frac{2}{3}}[(1+x)^{\frac{3}{2}}+x^{\frac{3}{2}}]+c}$