Question

Find the integrals (primitives):
$\rm \displaystyle\int \frac{2\ dx}{\sqrt{x}-\sqrt{x+3}}$

Answer 1

To solve the given integration we first simplify the expression by rationalising the denominator
$\frac{2}{\sqrt{x}-\sqrt{x+3}} \times \frac{ \sqrt{x} + \sqrt{x + 3} }{ \sqrt{x} + \sqrt{x + 3} } \\ \\ = \frac{2( \sqrt{x} + \sqrt{x + 3} )}{( { \sqrt{x} )}^{2} - ( { \sqrt{x + 3} })^{2} } \\ \\ = \frac{2( \sqrt{x} + \sqrt{x + 3}) }{ - 3} \\ \\ = \frac{ - 2}{3} \int \sqrt{x} dx - \frac{2}{3} \int \sqrt{x + 3} dx \\ \\ = \frac{ - 2}{3} \frac{ {x}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{2}{3} \frac{ {(x + 3)}^{ \frac{3}{2} } }{ \frac{3}{2} } + C\\ \\ = \frac{ - 4}{9} {x}^{ \frac{3}{2}} - \frac{ 4}{9} {(x + 3)}^{ \frac{3}{2}} + C \\ \\ \int \frac{2\ dx}{\sqrt{x}-\sqrt{x+3}}= \frac{ - 4}{9} \bigg( {x}^{ \frac{3}{2} } + {(x + 3)}^{ \frac{3}{2} }\bigg) + C$

Hope it helps you.