How do you integrate #int (x+4)/(x^2 + 2x + 5)dx# using partial fractions?

Question

Christopher Allers · Accepted Answer

#1/2[ln|x^2+2x+5|+3arctan((x+1)/2)]+c#

First split the integral into two parts:

#x+1# is a scalable factor of the derivative of #x^2+2x+5#, so divide #x+4# by #x+1#

#int(x+4)/(x^2+2x+5)dx=int(x+1)/(x^2+2x+5)dx+int3/(x^2+2x+5)dx#

Algebraic manipulation yields: #1/2int(2x+2)/(x^2+2x+5)dx+3int1/((x+1)^2+4)dx# #=1/2ln|x^2+2x+5|+3/2arctan((x+1)/2)+c# #=1/2[ln|x^2+2x+5|+3arctan((x+1)/2)]+c#

Adam Adkins · Answer

undefined To integrate $ \int \frac{x+4}{x^2 + 2x + 5} \, dx $ using partial fractions, we first decompose the fraction into partial fractions.

Given $ \frac{x+4}{x^2 + 2x + 5} $, we can express it as $ \frac{A}{x^2 + 2x + 5} + \frac{B}{x^2 + 2x + 5} $, where $ A $ and $ B $ are constants to be determined.

Next, we need to find $ A $ and $ B $ by equating the numerator of the original fraction to the sum of the numerators of the partial fractions:

$$ x + 4 = A(x^2 + 2x + 5) + B(x + 2) $$

Expanding and collecting like terms, we get:

$$ x + 4 = (A + B) x^2 + (2A + B) x + 5A + 2B $$

Comparing coefficients, we have:

$$ A + B = 0 $$
$$ 2A + B = 1 $$
$$ 5A + 2B = 4 $$

Solving this system of equations yields $ A = 1 $ and $ B = -1 $.

Now, we rewrite the integral using the partial fractions:

$$ \int \frac{x+4}{x^2 + 2x + 5} \, dx = \int \frac{1}{x^2 + 2x + 5} \, dx - \int \frac{1}{x^2 + 2x + 5} \, dx $$

The integrals on the right-hand side can be evaluated using the arctangent function and the natural logarithm.

Thus,

$$ \int \frac{x+4}{x^2 + 2x + 5} \, dx = \arctan\left(\frac{x+1}{2}\right) - \ln|x^2 + 2x + 5| + C $$

Where $ C $ is the constant of integration.