# How do you find the integral #int sqrt(4x-5)dx# using substitution?

##### 1 Answer

Feb 5, 2018

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

#### Explanation:

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 #