# Math Help - Integral with complex denominator

1. ## Integral with complex denominator

$\int\frac{8x+3}{4x^2+12x+11}dx$

I am not able to solve it using partial fractions because denominator doesn't have real values. What should I do ?

2. Originally Posted by totalnewbie
$\int\frac{8x+3}{4x^2+12x+11}dx$

I am not able to solve it using partial fractions because denominator doesn't have real values. What should I do ?
In that case you write it as,

$\frac{8x+3}{4x^2+12x+11}=\frac{Ax+B}{4x^2+12x+11}+ \frac{Cx+D}{4x^2+12x+11}$

3. Originally Posted by ThePerfectHacker
In that case you write it as,

$\frac{8x+3}{4x^2+12x+11}=\frac{Ax+B}{4x^2+12x+11}+ \frac{Cx+D}{4x^2+12x+11}$
Does it work out ?
$\int\frac{8x+3}{4x^2+12x+11}dx=\frac{1}{2}\int\fra c{8x+12}{4x^2+12x+11}dx-\int\frac{8}{4x^2+12x+11}dx$
Is my approach right ? I am not sure if constant $\frac{1}{2}$ is right.

4. Originally Posted by totalnewbie
Does it work out ?
$\int\frac{8x+3}{4x^2+12x+11}dx=\frac{1}{2}\int\fra c{8x+12}{4x^2+12x+11}dx-\int\frac{8}{4x^2+12x+11}dx$
Is my approach right ? I am not sure if constant $\frac{1}{2}$ is right.
I do not think so, with or without the konstant.

5. Hello, totalnewbie!

$\int\frac{8x+3}{4x^2+12x+11}\,dx$

We have: . $\int\frac{8x + 12 - 9}{4x^2+12x+11}\,dx \:=\:\int\frac{8x + 12}{4x^2+12x+11}\,dx \:- 9\int\frac{dx}{4x^2+12x+11}$

The first integral is of the form: . $\int\frac{du}{u}$

For the second integral, complete the square:
. . $4\left(x^2 + 3x + \frac{11}{4}\right) \;= \;4\left(x^2 + 3x + \frac{9}{4} + \frac{11}{4} - \frac{9}{4}\right) \;=\;4\left(\left[x+\frac{3}{2}\right]^2 + \frac{1}{2}\right)$

The integral becomes: . $-\frac{9}{4}\int\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}$

Trig Substitution: .Let $x + \frac{3}{2}\:=\:\frac{1}{\sqrt{2}}\tan\theta$ . . . etc.