Student Question

# int sqrt(x)arctan(x^(3/2)) dx Use integration tables to find the indefinite integral.

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)