Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 30

For the given integral problem: int sqrt(x)arctan(x^(3/2))dx , we can evaluate this applying indefinite integral formula: 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.
From the basic indefinite integration table, the problem resembles one of the formula for integral of inverse trigonometric function:
int arctan(u) du = u * arctan(u)- ln(u^2+1)/2+C
For easier comparison, we may apply u-substitution by letting:
u =x^(3/2 )
To determine the derivative of u, we apply the Power rule for derivative:d/(dx) x^n = n*x^(n-1) dx.
du =d/(dx) x^(3/2)
= (3/2) *x^(3/2-1) * 1 dx
= 3/2x^(1/2) dx
=3/2sqrt(x) dx
Rearrange du =3/2sqrt(x) dx into (2du)/3 = sqrt(x) dx .
Plug-in the values u = x^3/2 and (2du)/3 = sqrt(x) dx , we get:
int sqrt(x)arctan(x^(3/2))dx =int arctan(x^(3/2))*sqrt(x)dx
= int arctan(u) *(2du)/3
Apply the basic integration property: int c*f(x) dx = c int f(x) dx .
int arctan(u) *(2du)/3 =2/3int arctan(u)du.
Applying the aforementioned formula for inverse trigonometric function, we get:
2/3int arctan(u)du=(2/3) *[u * arctan(u)- ln(u^2+1)/2]+C
=(2u * arctan(u))/3- (2ln(u^2+1))/6+C
=(2u * arctan(u))/3- ln(u^2+1)/3+C
Plug-in u =x^(3/2) on (2u * arctan(u))/3- ln(u^2+1)/3+C , we get the indefinite integral as:
int sqrt(x)arctan(x^(3/2))dx =(2x^(3/2) * arctan(x^(3/2)))/3- ln((x^(3/2))^2+1)/3+C
=(2x^(3/2) * arctan(x^(3/2)))/3- ln(x^3+1)/3+C
or (2xsqrt(x) arctan(xsqrt(x)))/3- ln(x^3+1)/3+C

Note: x^(3/2) = xsqrt(x)