# Did I do this properly? Integration by Partial Fractions

1. Mar 16, 2008

### the7joker7

1. The problem statement, all variables and given/known data

Evaluate the indefinite integral.

int (6 x + 7)/(x^2 + 1) dx `

3. The attempt at a solution

A/(x + 1) + B/(x - 1)

6x + 7 = A(x - 1) + B(x + 1)

6x + 7 = (A + B)x + (-A + B)

A + B = 6

-A + B = 7

A + (7 + A) = 6

2A = -1.

A = -.5

B = 3.5

So the answer should be...

-.5ln(x + 1) + 3.5ln(x - 1) + C

Is this correct?

2. Mar 16, 2008

### rocomath

Differentiate your anti-derivative and see if it becomes your original Integral.

Also, I don't think it is. How did you break ... $$x^2+1$$ ???

You don't need to use Partial Fractions.

$$\int\frac{6x+7}{x^2+1}dx$$

$$\int\left(\frac{6x}{x^2+1}+\frac{7}{x^2+1}\right)dx$$

Last edited: Mar 16, 2008
3. Mar 16, 2008

### the7joker7

-.5(1/x + 1) + 3.5(1/x - 1)

Blah. =/

What did I do wrong?

4. Mar 16, 2008

### rocomath

Looking at your work ... looks like you broke $$x^2+1$$ to $$(x+1)(x-1)$$

Yes?

$$x^2-1=(x+1)(x-1)$$ so ... $$x^2+1\neq(x+1)(x-1)$$