How do you integrate #(5x+1)/((x+3)(x+2)(x-4))# using partial fractions?

Question

Aurora Alvarado · Accepted Answer

#int ((5x+1)*dx)/[(x+3)(x+2)(x-4)]#

=#1/2*ln(x-4)+3/2*ln(x+2)-2ln(x+3)+C# I decomposed integrand into basic fractions, #(5x+1)/[(x+3)(x+2)(x-4)]# =#A/(x+3)+B/(x+2)+C/(x-4)# After expanding denominator, #A(x+2)(x-4)+B(x+3)(x-4)+C(x+3)(x+2)=5x+1# After setting #x=-3#, #7A=-14#, so #A=-2# After setting #x=-2#, #-6B=-9#, so #B=3/2# After setting #x=4#, #42C=21#, so #C=1/2# Thus, #int ((5x+1)*dx)/[(x+3)(x+2)(x-4)]# =#-2*int (dx)/(x+3)+3/2*int (dx)/(x+2)+1/2*int dx/(x-4)# =#1/2*ln(x-4)+3/2*ln(x+2)-2ln(x+3)+C#

Charles Arocha · Answer

undefined To integrate $(5x+1)/((x+3)(x+2)(x-4))$ using partial fractions, you first need to express the rational function as a sum of simpler fractions. The partial fraction decomposition would be:

$$\frac{5x+1}{(x+3)(x+2)(x-4)} = \frac{A}{x+3} + \frac{B}{x+2} + \frac{C}{x-4}$$

To find the constants $A$, $B$, and $C$, you can multiply both sides by $(x+3)(x+2)(x-4)$ to clear the denominators:

$$5x + 1 = A(x+2)(x-4) + B(x+3)(x-4) + C(x+3)(x+2)$$

Then, you can solve for $A$, $B$, and $C$ by choosing appropriate values of $x$ to eliminate the unknowns.

After finding $A$, $B$, and $C$, you integrate each term separately. The integral of $\frac{A}{x+3}$ would be $A\ln|x+3|$, the integral of $\frac{B}{x+2}$ would be $B\ln|x+2|$, and the integral of $\frac{C}{x-4}$ would be $C\ln|x-4|$.

Therefore, the integral of $(5x+1)/((x+3)(x+2)(x-4))$ using partial fractions would be:

$$A\ln|x+3| + B\ln|x+2| + C\ln|x-4| + \text{constant}$$

Where $A$, $B$, and $C$ are the constants obtained from the partial fraction decomposition.