Question

How do you find the integral of `int (x-2)/((x+1)^2+4)`?

Answer 1

`1/2ln(x^2+2x+5)-3(1/2arctan((x+1)/2))+C`

Explanation:

`int(x-2)/((x+1)^2+4)dx`
you can't use the integral of arctan(x) yet, because the numerator isn't just a constant. to remove the x term from the numerator, you have to use ln(x).

the formula for using ln(x) for integrals: `int(f'(x))/f(x)dx=ln(f(x))+C`

expand the denominator: `int(x-2)/(x^2+2x+5)dx`. now you want to change the numerator to be the derivative of the denominator `x^2+2x+5`. that means you want to create two fractions, one of which has the numerator `d/dx(x^2+2x+5)` or `2x+2`

rewrite `int(x-2)/(x^2+2x+5)dx` as `int(x+1-3)/(x^2+2x+5)dx = int(x+1)/(x^2+2x+5)dx - int3/(x^2+2x+5)dx = 1/2int(2x+2)/(x^2+2x+5)dx - int3/((x+1)^2+4)dx`

now the first integral has the `(f'(x))/f(x)` format and the second integral will now integrate into some form of `arctan(x)`

`1/2int(2x+2)/(x^2+2x+5)dx =1/2ln(x^2+2x+5)+C`

for `- int3/((x+1)^2+4)dx`, use the formula: ![http://www.dummies.com/education/math/calculus/integrating-using-partial-fractions-when-the-denominator-contains-irreducible-quadratic-factors/]()

for this problem, (x+1) will equal the x in the formula, and 2 will equal the a.
`-3int1/((x+1)^2+4)dx = -3(1/2arctan((x+1)/2))+C`

combining everything:
`int(x-2)/((x+1)^2+4)dx = 1/2ln(x^2+2x+5)+C-3(1/2arctan((x+1)/2))+C`
you can merge the two C's into a single C because adding any two constants still results in a constant.

final answer: `1/2ln(x^2+2x+5)-3(1/2arctan((x+1)/2))+C`