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#
