How do you integrate #int5xsqrt(2x+3)# using substitution?

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Answer 1 · 2016-07-17T11:16:55

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

This integral doesn't seem quiet obvious, so I will go through all the steps in high detail.

When evaluating integrals, it is sometimes a matter of trial and error, especially when it comes to u-substitutions.

For this particular integral, we may choose #u = 2x+3# and see what happens.

If #u=2x+3#, we'd have a #du# floating around somewhere - but there isn't. However, since there's a #5x# out front, we may be able to solve for #x#.

#u = 2x+3 -> du = 2 dx -> 1/2 du = dx#

Solving #u# for #x# we get

#u-3 = 2x -> (u-3)/(2) = x#

We can, however, pull out a #1/2# again, since it's just a constant.

Proceeding with the integration process, we can now rewrite our integral.

#int 5x * sqrt(2x+3) dx = 5 * 1/2 * 1/2 int (u-3) u^(1/2) du#

If we multiply everything into the parentheses and simplify we get

#5/4 int u^(3/2) - 3u^(1/2) du = 5/4[ 2/5u^(5/2) - 2/cancel(3) * cancel(3) u^(3/2)]+C#

#=1/2u^(5/2) - 5/2 u^(3/2)+C#

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