`int x^2sqrt(2+9x^2) dx` Use integration tables to find the indefinite integral.

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marizi eNotes educator| Certified Educator

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`