Question

evaluate indefinite integral

Answer 1

here is your answer...........

Answer 2

$\displaystyle \int \sf \small\frac{dx}{ \sqrt{x + 3} + \sqrt{x + 2} }$

$\displaystyle = \int \sf \small\frac{1 \times ( \sqrt{x + 3} - \sqrt{x + 2}) }{ (\sqrt{x + 3} + \sqrt{x + 2}) ( \sqrt{x + 3} - \sqrt{x + 2} )} \: dx$

$\displaystyle = \int \sf \small \frac{ \sqrt{x + 3} + \sqrt{x + 2} }{x + 3 - x - 2} \: dx$

$\displaystyle = \int \sf \small \frac{ \sqrt{x + 3} + \sqrt{x + 2} }{1} \: dx$

$\displaystyle = \int \sf \small (\sqrt{x + 3} + \sqrt{x + 2}) \: dx$

$\displaystyle = \small\int \sf \sqrt{x + 3} \: dx+ \int \sqrt{x + 2} \: dx$

Using $\sf {x}^{n} dx = \frac{ {x}^{n + 1} }{n + 1} + C$We get,

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

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

$\sf = \frac{ 2{(x + 3)}^{ \frac{3}{2} } }{3} + \frac{ 2{(x + 2)}^{ \frac{3}{2} } }{3 }+C$

This is the required answer.