How do you use partial fractions to find the integral int (x^2-1)/(x^3+x)dx∫x2−1x3+xdx?
1 Answer
Feb 28, 2017
Explanation:
Start by factoring the denominator.
x^3 + x = x(x^2 + 1)x3+x=x(x2+1)
Now write the partial fraction decomposition.
(Ax + B)/(x^2 + 1) + C/x = (x^2 - 1)/(x(x^2 + 1))Ax+Bx2+1+Cx=x2−1x(x2+1)
(Ax + B)x + C(x^2 + 1) = x^2 - 1(Ax+B)x+C(x2+1)=x2−1
Ax^2 + Bx + Cx^2 + C = x^2 - 1Ax2+Bx+Cx2+C=x2−1
(A + C)x^2 + Bx + C = x^2 - 1(A+C)x2+Bx+C=x2−1
We now write a system of equations.
{(A + C = 1), (B = 0), (C = -1):}
Solving, we get
int(2x)/(x^2 + 1) - 1/xdx
int (2x)/(x^2 + 1)dx - int 1/xdx
We now make the substitution
int (2x)/u * (du)/(2x) - int 1/xdx
int 1/u du - int 1/x dx
ln|u| - ln|x| + C
ln|x^2 + 1| - ln|x| + C
ln|(x^2 + 1)/x|+ C
Hopefully this helps!
