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

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Answer 1 · 2018-02-05T01:36:57

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

We seek:

$I = int \ sqrt(4x-5) \ dx$

We can perform a substitution:

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

Then, we can substitute into the integral, as follows:

$I = 1/4 \ int \ sqrt(4x-5) \ (4) \ dx$
$\ \ = 1/4 \ int \ sqrt(u) \ du$
$\ \ = 1/4 \ int \ u^(1/2) \ du$

Which is a standard integral, so using the power rule:

$I = 1/4 \ u^(3/2) / (3/2) + C$
$\ \ = 1/4 \ 2/3 \ u^(3/2) + C$
$\ \ = 1/6 \ u^(3/2) + C$

And then restoring the substitution, we find that:

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

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