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

1 Answer
Mar 3, 2018

#intsqrt(x^3+x^2)(3x^2+2x)dx=2/3(x^3+x^2)^(3/2)+C#

Explanation:

Let's set #u=x^3+x^2.#

Differentiating, we get

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

We can change this derivative to a differential by moving #dx# to the right side (multiplying both sides by it)

#du=(3x^2+2x)dx#

Look at our integral, #intsqrt(x^3+x^2)(3x^2+2x)dx#

We see #(3x^2+2x)dx# appears in the integral; thus, we can replace it with #du.# Furthermore, we already said #u=x^3+x^2,# so we can replace everything under the root with #u.#

#intsqrt(x^3+x^2)(3x^2+2x)dx=intsqrt(u)du#

Let's rewrite this with exponents:

#intsqrt(u)du=intu^(1/2)du#

Recall that #intx^adx=x^(a+1)/(a+1)+C#; where #C# is just the constant of integration; therefore,

#intu^(1/2)du=u^(1/2+1)/(1/2+1)+C=u^(3/2)/(3/2)+C=2/3u^(3/2)+C#

Let's rewrite in terms of #x,# recalling that #u=x^3+x^2:#

#2/3u^(3/2)+C=2/3(x^3+x^2)^(3/2)+C#