How do you use partial fractions to find the integral #int (x+1)/(x^2+4x+3) dx#?

Question

Evelyn Agard · Accepted Answer

#int(x+1)/(x^2+4x+3)dx#

We can begin by factoring the denominator of the integrand:

#=>int(x+1)/((x+1)(x+3))dx#

Cancel like terms:

#=>intcancel((x+1))/(cancel((x+1))(x+3))dx#

#=>int1/(x+3)dx#

We can now use a simple substitution:

#u=x+3, du=dx#

#=>int 1/udu#

#=>ln(|u|)+C#

Substituting back in:

#=>ln(|x+3|)+C#

Isabella Ackerman · Answer

undefined To use partial fractions to find the integral of $ \int \frac{x + 1}{x^2 + 4x + 3} \, dx $, follow these steps:

1. Factor the denominator $ x^2 + 4x + 3 $ into irreducible factors.
2. Express the fraction $ \frac{x + 1}{x^2 + 4x + 3} $ as a sum of partial fractions with undetermined constants.
3. Solve for the undetermined constants by equating coefficients.
4. Integrate each partial fraction individually.
5. Combine the results to obtain the final integral.

The steps are as follows:

1. Factor the denominator: $ x^2 + 4x + 3 = (x + 3)(x + 1) $.

2. Express the fraction as partial fractions: $ \frac{x + 1}{x^2 + 4x + 3} = \frac{A}{x + 3} + \frac{B}{x + 1} $.

3. Solve for $ A $ and $ B $:

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

Expand and equate coefficients:

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

Equating coefficients of like terms:

$ A + B = 1 $
   $ A + 3B = 1 $

Solve this system of equations to find $ A $ and $ B $.

4. Once you find the values of $ A $ and $ B $, integrate each partial fraction separately:

$ \int \frac{A}{x + 3} \, dx $ and $ \int \frac{B}{x + 1} \, dx $.

5. Combine the results to obtain the final integral:

$ \int \frac{x + 1}{x^2 + 4x + 3} \, dx = A \ln|x + 3| + B \ln|x + 1| + C $, where $ C $ is the constant of integration.