Asked by alice
Evaluate the integral. (Remember to use absolute values where appropriate. Use C for the constant of integration.)
(x^2−x+12)/(x^3+3x) dx
(x^2−x+12)/(x^3+3x) dx
Answers
Answered by
Steve
Using partial fractions,
(x^2-x+12)/(x^3+3x)
= 4/x - (3x+1)/(x^2+3)
The 4/x integrates easily enough. The rest has to be worked on...
3x/(x^2+3) is just another log. For the rest let
x = ?3 tan?
x^2+3 = 3 sec^2?
dx = ?3 sec^2? d?
1/(x^2+3) dx = 1/?3 d?
Now, putting all that together, you get
http://www.wolframalpha.com/input/?i=integral+(x%5E2-x%2B12)%2F(x%5E3%2B3x)
(x^2-x+12)/(x^3+3x)
= 4/x - (3x+1)/(x^2+3)
The 4/x integrates easily enough. The rest has to be worked on...
3x/(x^2+3) is just another log. For the rest let
x = ?3 tan?
x^2+3 = 3 sec^2?
dx = ?3 sec^2? d?
1/(x^2+3) dx = 1/?3 d?
Now, putting all that together, you get
http://www.wolframalpha.com/input/?i=integral+(x%5E2-x%2B12)%2F(x%5E3%2B3x)
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