Question

Question #2a34f

Answer 1

`intx^2/(9+16x^6)dx = 1/36arctan(4/3x^3)+"constant"`

Explanation:

For the integral `F(x)=intx^2/(9+16x^6)dx`, sub `u=x^3` and `du=3x^2dx`

Then `F(x) = 1/3int1/(9+16u^2)du = 1/27int1/(1+16/9u^2)du`

For the integral `F(x)`, sub `t=4/3u` and `dt=4/3du`

Then `F(x) = 1/36int1/(1+t^2)dt`

`F(x)` is now in a standard form and can be evaluated as a standard integral.

`F(x)=1/36int1/(1+t^2)dt=1/36arctant = 1/36arctan(4/3u) = 1/36arctan(4/3x^3)+"constant"`

Answer 2

`1/36tan^-1(4/3x^3)+C`

Explanation:

`int(x^2dx)/(9+16x^6)`

Use the substitution `x^3=3/4tantheta,=>3x^2dx=3/4sec^2thetad theta`.

`=1/3int(3x^2dx)/(9+16(x^3)^2)=1/3int(3/4sec^2thetad theta)/(9+16(9/16tan^2theta))`

`=1/4int(sec^2thetad theta)/(9(1+tan^2theta))`

Recall that `1+tan^2theta=sec^2theta`:

`=1/36intd theta=1/36theta+C`

From `x^3=3/4tantheta`, note that `tantheta=4/3x^3` so `theta=tan^-1(4/3x^3)`:

`=1/36tan^-1(4/3x^3)+C`