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

Question

Charles Arocha · Accepted Answer

The answer is #=5/4ln(x-1)+15/4ln(x+3)+C# First, let's do the decomposition into partial fractions, #(5x)/((x-1)(x+3))=A/(x-1)+B/(x+3)# #=((x+3)+B(x-1))/((x-1)(x+3))# #:. 5x=A(x+3)+B(x-1)# Let #x=-3##=>##-15=-4B##=>##B=15/4# Let #x=1##=>##5=4A##=>##A=5/4# #(5x)/((x-1)(x+3))=(5/4)/(x-1)+(15/4)/(x+3)# So, #int(5xdx)/((x-1)(x+3))=int(5/4dx)/(x-1)+int(15/4dx)/(x+3)# #=5/4ln(x-1)+15/4ln(x+3)+C#

Charles Arocha · Answer

undefined To integrate $\frac{5x}{(x-1)(x+3)}$ using partial fractions, we first express the fraction as the sum of two simpler fractions. We write:

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

Next, we clear the denominators by multiplying both sides of the equation by $(x-1)(x+3)$. This gives us:

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

Expanding and collecting like terms:

$$5x = Ax + 3A + Bx - B$$

Now, we equate coefficients of like terms:

For $x$: $A + B = 5$

For constants: $3A - B = 0$

Solving this system of equations, we find $A = 2$ and $B = 3$.

Therefore, we can express the original fraction as:

$$\frac{5x}{(x-1)(x+3)} = \frac{2}{x-1} + \frac{3}{x+3}$$

Now, we integrate each term separately:

$$\int \frac{5x}{(x-1)(x+3)} \, dx = 2\int \frac{1}{x-1} \, dx + 3\int \frac{1}{x+3} \, dx$$

This gives us:

$$= 2\ln|x-1| + 3\ln|x+3| + C$$

Where $C$ is the constant of integration.