1. ## Partial Fractions

If anyone could answer a question, I would appreciate it.

I'm working on solving $\displaystyle \int\frac{5-x}{2x^2+x-1} dx$ using partial fractions.

Here's a summary of my work: $\displaystyle \frac{5-x}{2x^2+x-1} = \frac{A}{2x-1}+\frac{B}{x+1}$ which leads to $\displaystyle 5-x=A(x+1)+B(2x-1)$.

Letting x = -1 yields -3B = 6, B = -2. Letting x = 1/2 yields 9/2 = 3/2A, or A = 3. This gives an integral with partial fractions: $\displaystyle \int(\frac{3}{2x-1}-\frac{2}{x+1}) dx($ which gives 3ln|2x-1|-2ln|x+1| + C.

The text answer is (3/2)ln|2x-1|-2ln|x+1| + C. What am I missing?

2. Originally Posted by kaiser0792
If anyone could answer a question, I would appreciate it.

I'm working on solving $\displaystyle \int\frac{5-x}{2x^2+x-1} dx$ using partial fractions.

Here's a summary of my work: $\displaystyle \frac{5-x}{2x^2+x-1} = \frac{A}{2x-1}+\frac{B}{x+1}$ which leads to $\displaystyle 5-x=A(x+1)+B(2x-1)$.

Letting x = -1 yields -3B = 6, B = -2. Letting x = 1/2 yields 9/2 = 3/2A, or A = 3. This gives an integral with partial fractions: $\displaystyle \int(\frac{3}{2x-1}-\frac{2}{x+1}) dx($ which gives 3ln|2x-1|-2ln|x+1| + C.

The text answer is (3/2)ln|2x-1|-2ln|x+1| + C. What am I missing?

Your partial fractions are OK but be caution :
The integral $\displaystyle \int \frac{dx}{2x-1}$ is $\displaystyle \frac{1}{2}\ln|2x-1| + C$ . You have missed the $\displaystyle \frac{1}{2}$ .

3. Thanks simplependulum, a careless mistake on my part, but your insight saved me much frustration.