How do you find the integral #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 #