$\displaystyle \int\frac{8x+3}{4x^2+12x+11}dx$
I am not able to solve it using partial fractions because denominator doesn't have real values. What should I do ?
Hello, totalnewbie!
$\displaystyle \int\frac{8x+3}{4x^2+12x+11}\,dx$
We have: .$\displaystyle \int\frac{8x + 12 - 9}{4x^2+12x+11}\,dx \:=\:\int\frac{8x + 12}{4x^2+12x+11}\,dx \:- 9\int\frac{dx}{4x^2+12x+11}$
The first integral is of the form: .$\displaystyle \int\frac{du}{u}$
For the second integral, complete the square:
. . $\displaystyle 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: .$\displaystyle -\frac{9}{4}\int\frac{dx}{\left(x + \frac{3}{2}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} $
Trig Substitution: .Let $\displaystyle x + \frac{3}{2}\:=\:\frac{1}{\sqrt{2}}\tan\theta$ . . . etc.