x^3-x^2+x+3=(x+1)(x^2-2x+3)
(x^2+5)/(x^3-x^2+x+3)=
=A/(x+1)+(Bx+C)/(x^2-2x+3)
A(x^2-2x+3)+(Bx+C)(x+1)=x^2+5
A+B=1
-2A+B+C=0
3A+C=5
Find A,B,C
Please help me integrate this equation using partial fractions:
Integrate [(x^2+5)/(x^3-x^2+x+3)]dx.
Thank you very much.
3 answers
In google type:
wolfram alpha
When you see list of results click on:
Wolfram Alpha:Computational Knoweledge Engine
When page be open in rectangle type:
(x^2+5)/(x^3-x^2+x+3) and click option =
After few secons you will see all about that function.
Now click option:
Partial fraction expansion: Show steps
Now in wolfram alpha rectangle type:
2/(x^2-2x+3)+1/(x+1)
and click option =
When you see results clic option:
Indefinite integral: Show steps
wolfram alpha
When you see list of results click on:
Wolfram Alpha:Computational Knoweledge Engine
When page be open in rectangle type:
(x^2+5)/(x^3-x^2+x+3) and click option =
After few secons you will see all about that function.
Now click option:
Partial fraction expansion: Show steps
Now in wolfram alpha rectangle type:
2/(x^2-2x+3)+1/(x+1)
and click option =
When you see results clic option:
Indefinite integral: Show steps
Thank you very much.