How do you evaluate the integral $int sqrt(2x+3)dx$ ?

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Answer 1 · 2017-01-09T03:46:21

$= 1/3(2x + 3)^(3/2)+ C$

Let $u = 2x + 3$ . Then $du = 2dx$ and $dx= 1/2du$

$=1/2intsqrt(u)du$

$=1/2int u^(1/2)du$

You can integrate this as $int(x^n)dx = x^(n + 1)/(n + 1) + C$ :

$=1/2(2/3u^(3/2)) + C$

$=1/3u^(3/2)+ C$

$= 1/3(2x + 3)^(3/2)+ C$

Hopefully this helps!

How do you evaluate the integral int sqrt(2x+3)dxint sqrt(2x+3)dx?