Answer:
[tex]\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx = \frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}(\frac{x}{\sqrt 3} )+ c[/tex]
Step-by-step explanation:
Given
[tex]\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx[/tex]
Required
Integrate
Using partial fraction, we have:
[tex]\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{Ax+B}{x^2 + 1} + \frac{Cx + D}{x^2 + 3}[/tex]
Take LCM
[tex]\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{(Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1)}{(x^2 + 1)(x^2 + 3)}[/tex]
Cancel out the denominators
[tex]3x^3 + 4x^2 + 9x + 4} = (Ax+B)(x^2 + 3)+ (Cx + D)(x^2 + 1)[/tex]
Open brackets
[tex]3x^3 + 4x^2 + 9x + 4} = Ax^3+Bx^2 + 3Ax +3B+ Cx^3 + Dx^2 + Cx + D[/tex]
Collect like terms
[tex]3x^3 + 4x^2 + 9x + 4 = Ax^3+ Cx^3+Bx^2+ Dx^2 + 3Ax+ Cx +3B + D[/tex]
Compare like terms on opposite sides
[tex]Ax^3 + Cx^3 = 3x^3[/tex] [tex]A + C = 3[/tex]
[tex]Bx^2 + Dx^2 = 4x^2[/tex] [tex]B + D = 4[/tex]
[tex]3Ax + Cx = 9x[/tex] [tex]3A + C = 9[/tex]
[tex]3B + D = 4[/tex]
Subtract [tex]B + D = 4[/tex] from [tex]3B + D = 4[/tex]
[tex]3B - B + D - D = 4 - 4[/tex]
[tex]2B + 0 = 0[/tex]
[tex]2B = 0[/tex]
[tex]B = 0[/tex]
[tex]B + D = 4[/tex]
[tex]D =4 - B[/tex]
[tex]D =4 - 0[/tex]
[tex]D =4[/tex]
Subtract [tex]A + C = 3[/tex] from [tex]3A + C = 9[/tex]
[tex]3A - A + C - C = 9 - 3[/tex]
[tex]2A = 6[/tex]
[tex]A = 3[/tex]
[tex]A + C = 3[/tex]
[tex]C = 3 - A[/tex]
[tex]C = 3 - 3[/tex]
[tex]C = 0[/tex]
So, we have:
[tex]A = 3[/tex] [tex]B = 0[/tex] [tex]C = 0[/tex] [tex]D =4[/tex]
[tex]\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{Ax+B}{x^2 + 1} + \frac{Cx + D}{x^2 + 3}[/tex]
[tex]\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{3x+0}{x^2 + 1} + \frac{0*x + 4}{x^2 + 3}[/tex]
[tex]\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} = \frac{3x}{x^2 + 1} + \frac{4}{x^2 + 3}[/tex]
The integral becomes:
[tex]\int\limits {[\frac{3x}{x^2 + 1} + \frac{4}{x^2 + 3}]} \, dx[/tex]
Split:
[tex]\int\limits {\frac{3x}{x^2 + 1} \, dx + \int\limits {\frac{4}{x^2 + 3}} \, dx[/tex]
Split
[tex]\frac{3}{2} \int\limits {\frac{2x}{x^2 + 1} \, dx + 4\int\limits {\frac{1}{x^2 + 3}} \ dx[/tex]
Integrate
[tex]\frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}\frac{x}{\sqrt 3} + c[/tex]
Hence:
[tex]\int\limits {\frac{3x^3 + 4x^2 + 9x + 4}{(x^2 + 1)(x^2 +3)}} \, dx = \frac{3}{2}\ln(x^2 + 1) + 4\sqrt{3}\tan^{-1}(\frac{x}{\sqrt 3} )+ c[/tex]
