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

1 Answer
May 3, 2016

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).