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

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)#.