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

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Answer 1 · 2016-07-18T15:42:35

Let $u = 4 x - 5$ $u = 4x - 5$ , then $\frac{d u}{d x} = 4 \to d x = \frac{d u}{4}$ $(du)/dx = 4 ->dx = (du)/4$

$= \frac{1}{4} \int (\sqrt{u})$ $=1/4int(sqrt(u))$

$= \frac{1}{4} (\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}) + C$ $=1/4(u^(1/2 + 1)/(1/2 + 1)) + C$

$= \frac{1}{4} (\frac{2}{3} u^{\frac{3}{2}}) + C$ $=1/4(2/3u^(3/2)) + C$

$= \frac{1}{4} (\frac{2}{3} {(4 x - 5)}^{\frac{3}{2}}) + C$ $=1/4(2/3(4x -5)^(3/2)) + C$

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

Hopefully this helps!

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