Partial Fractions

**kaiser0792** · April 11th 2010, 07:34 PM

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

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

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

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: $\int(\frac{3}{2x-1}-\frac{2}{x+1}) dx($ 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?

**simplependulum** · April 11th 2010, 07:41 PM

Originally Posted by kaiser0792

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

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

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

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: $\int(\frac{3}{2x-1}-\frac{2}{x+1}) dx($ 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 $\int \frac{dx}{2x-1}$ is $\frac{1}{2}\ln|2x-1| + C$ . You have missed the $\frac{1}{2}$ .

**kaiser0792** · April 11th 2010, 08:01 PM

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

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