How do you integrate $int sqrt(4x-5)dx$ using substitution?

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Answer 1 · 2016-07-18T15:42:35

Let $u = 4x - 5$ , then $(du)/dx = 4 ->dx = (du)/4$

$=1/4int(sqrt(u))$

$=1/4(u^(1/2 + 1)/(1/2 + 1)) + C$

$=1/4(2/3u^(3/2)) + C$

$=1/4(2/3(4x -5)^(3/2)) + C$

$=1/6(4x - 5)^(3/2) + C$

Hopefully this helps!

How do you integrate int sqrt(4x-5)dxint sqrt(4x-5)dx using substitution?