Integral 3x^2+8/x^3+4x by Partial Fraction Decomposition

In this video, I demonstrate how to integrate the quotient (3x² + 8) / [x(x² + 4)].

Now, because we have a quotient of 2 polynomials, that we will have to perform partial fraction decomposition first to turn this integral into a form that we can integrate.

The first partial fraction is going to be a constant over x, or A/x.

The second partial fraction, because the quadratic x² + 4 is irreducible, we have to leave it as it is. But because we have a quadratic denominator, we have to have a numerator that is one degree less than the denominator, ie. Bx+C / x²+4.

And thus...

\frac{3x^2+8}{x\left ( x^2+4 \right )} = \frac{A}{x}+\frac{Bx+C}{x^2+4}

Now bringing the partial fractions back together through cross multiplication, we get...

\frac{3x^2+8}{x\left ( x^2+4 \right )} = \frac{A\left ( x^2+4 \right )+\left ( Bx+C \right )x}{x\left ( x^2+4 \right )}

Therefore...

3x^2+8 = \left ( A + B \right )x^2 + Cx + 4A

And now matching the coefficients to solve for the unknowns A, B and C, we get...

A = 2
B = 1
C = 0

Thus...

\int \frac{3x^2+8}{x\left ( x^2+4 \right )}\mathrm{d}x = 2\int\frac{\mathrm{d}x}{x}+\int \frac{x}{x^2+4}\mathrm{d}x

And we can integrate the 2 components separately.