Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 28

Recall that indefinite integral follows int f(x) dx = F(x) +C
where:
f(x) as the integrand function
F(x) as the antiderivative of f(x)
C as the constant of integration.
The given integral problem: int x^2sqrt(2+9x^2) dx resembles one of the formulas from the integration table. We follow the integral formula for function with roots as:
int u^2sqrt(a^2+u^2)du = u/8(a^2+2u^2)sqrt(a^2+u^2) -a^4/8ln|u+sqrt(a^2+u^2)| +C .
For easier comparison, we apply u-substitution by letting:
u^2 = 9x^2 or (3x)^2
then u = 3x or x=u/3 .
For the derivative of u, we get: du = 3 dx or (du)/3 = dx .
Note: The corresponding value of a^2=2
then a =sqrt(2) and a^4 = (a^2)^2 =2^2 =4
Plug-in the values of u = 3x , x=u/3 and (du)/3 = dx , we get:
int x^2sqrt(2+9x^2)dx=int (u/3)^2sqrt(2+u^2)* (du)/3
=int u^2/9*sqrt(2+u^2)* (du)/3
=int u^2/27sqrt(2+u^2)du
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int u^2/27*sqrt(2+u^2)*du =1/27int u^2sqrt(2+u^2)du
Apply the aforementioned integral formula with a^2 =2 , we get:
1/27 int u^2sqrt(2+u^2)du=1/27*[u/8(2+2u^2)sqrt(2+u^2) -4/8ln|u+sqrt(2+u^2)|]+C
= u/216(2+2u^2)sqrt(2+u^2) -1/2ln|u+sqrt(2+u^2)|+C
Plug-in u =3x on u/216(2+2u^2)sqrt(2+u^2) -1/2ln|u+sqrt(2+u^2)|+C , we get the indefinite integral as:
int x^2sqrt(2+9x^2) dx=(3x)/216(2+2(3x)^2)sqrt(2+(3x)^2) -1/2ln|3x+sqrt(2+(3x)^2)|+C
= x/72(2+18x^2)sqrt(2+9x^2) -1/2ln|3x+sqrt(2+9x^2)|+C
= ((2x+18x^3)sqrt(2+9x^2))/72 -(ln|3x+sqrt(2+9x^2)|)/2+C
or (xsqrt(2+9x^2))/36+(x^3sqrt(2+9x^2))/4 -(ln|3x+sqrt(2+9x^2)|)/2+C