How do you integrate #int (2x^(3/2)+1)sqrtx# using substitution?

1 Answer
Mar 29, 2018

#int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C#

Explanation:

Our goal with substitution is to find an expression whose differential appears in the integral, and represent it using the variable #u.#

We have #int(2x^(3/2)+1)sqrtxdx#

In this case, #u=2x^(3/2)+1# is the best possible choice. Were we to choose #u=sqrtx, du=1/(2sqrtx)dx#, and this differential is not even close to anything in the integral.

Calculate its differential:

#du=(2)(3/2)x^(1/2)dx#

#du=2sqrtxdx#

Divide both sides by #2:#

#1/2du=sqrtxdx#

And we see #sqrtxdx# does in fact show up in the integral. So, rewrite with the substitution:

#1/2intudu=1/2(1/2)u^2+C=1/4u^2+C#

Rewriting in terms of #x# yields

#int(2x^(3/2)+1)sqrtxdx=1/4(2x^(3/2)+1)^2+C#