Question

evaluate the integral: int: [(x^3 + 3x^2 + x + 9)/(x^2 + 1)(x^2 + 3) ] dx

using the linear quotient rule I have:

let x^3 + 3x^2 + x + 9 = Ax+B (x^2 + 3) + Cx + D(x^2 + 1)

I am trying to solve for A B C and D...

Answer 1

You have to us the method of partial fractions:
\[

\frac{x^3+3 x^2+x+9}{\left(x^2+1\right) \left(x^2+3\right)}=\frac{A
x+B}{x^2+1}+\frac{C x+D}{x^2+3}
\]
After finding A, B ,C, D, the integral would become easy.

Answer 2

Use the cover-up method.

Answer 3

You will have to the work. To check your answer,
A=0; B=3; C=1 and D=0

Answer 4

Your integral is now
\[
\int \frac{x^3+3 x^2+x+9}{\left(x^2+1\right) \left(x^2+3\right)} \,
dx=\int \frac{3}{x^2+1} \, dx+\int \frac{x}{x^2+3} \, dx=\\
3 \tan^{-1} x + \frac 1 2 \ln(x^2 +3) + K

\]

Answer 5

I got it. Thank you so much for your help.
A= 0
B = 3
C = 1
D = 0.

Answer 6

yw