Question

How do you find the definite integral of `x^2dx / (x^3 + 9)` from `[-1, 1]`?

Answer 1

`1/3ln(5/4)`

Explanation:

We want to find:

`int_(-1)^1x^2/(x^3+9)dx`

For this, we will use substitution. Let:

`u=x^3+9" "=>" "du=3x^2dx`

Note that we have a `x^2dx` term in the integral, but we need `3x^2dx`. Thus, multiply the interior of the integral by `3` and balance this by multiplying the exterior of the integral by `1//3`.

`int_(-1)^1x^2/(x^3+9)dx=1/3int_(-1)^1(3x^2)/(x^3+9)dx`

Before we substitute in our values for `u` and `du`, we will have to determine what the integral's new bounds will be. (This is necessary since we are shifting the integral from being in respect to `x`, as `dx`, to `u`, as in `du`.)

The new bounds can be determined by plugging in the current bounds, `-1` and `1`, into the expression we've defined to be `u`.

`u(-1)=(-1)^3+9=-1+9=8`

`u(1)=1^3+9=1+9=10`

Combining the values of `u` and `du` with the new bounds, we see that:

`1/3int_(-1)^1(3x^2)/(x^3+9)dx=1/3int_8^10(du)/u`

Note that `int(du)/u=lnabsu+C`.

`1/3int_8^10(du)/u=1/3[lnabsu]_8^10=1/3(ln10-ln8)`

There are various ways this can be simplified:

`1/3(ln10-ln8)=1/3ln(10/8)=1/3ln(5/4)=lnroot3(5/4)`

The best simplification (most standard) is likely `1/3ln(5/4)`.