Question

fractional decomposition

Answer 1

\[\int\limits \frac{ x^{3}-2x^{2}-3x+1 }{ x^{2}+4x+5 }dx\]

Answer 2

first do long division

Answer 3

I did divided the polynomials and got\[\int\limits x-6+\frac{ 16x+31 }{ x^{2}+4x+5 }\]

Answer 4

my problem is I dont know how to progress from here

Answer 5

@perl

Answer 6

ok so the difficulty is integrating (16x +31) / (x^2+4x + 5)

Answer 7

yes I tried to get in a form for arctan but splitting it after I didnt see what I could do still\[\int\limits x-6+\frac{ 16x+31 }{ (x+2)^2+1^2 }\]

Answer 8

we can rewrite the numerator

Answer 9

16x + 31 = 8(2x + 4) - 1

Answer 10

ok

Answer 11

and from there split it and u sub?

Answer 12

\[\int\limits\limits x-6+\frac{ 16x+31 }{ x^{2}+4x+5 } =\int\limits\limits x-6+\frac{ 8(2x+4) }{ x^{2}+4x+5 } -\frac{1}{x^{2}+4x+5} \]

Answer 13

right, and you can complete the square for the last term

Answer 14

?

Answer 15

oh ok. Thank you for all your help.

Answer 16

\[\int\limits\limits \frac{ x^{3}-2x^{2}-3x+1 }{ x^{2}+4x+5 }dx \\= \int\limits\limits x-6+\frac{ 16x+31 }{ x^{2}+4x+5 } \\ =\int\limits\limits x-6+\frac{ 8(2x+4)-1 }{ x^{2}+4x+5} \\=\int\limits\limits x-6+\frac{ 8(2x+4) }{ x^{2}+4x+5} -\frac{1}{x^{2}+4x+5} \\ =\int\limits\limits x-6+\frac{ 8(2x+4) }{ x^{2}+4x+5 } -\frac{1}{(x+2)^2+1} \]