Student Question

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

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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)`

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