How Do You Solve Complex Integrals Involving Rational Functions?

[luigihs]

Hello I am trying to integrate this problem:
∫x-1/x^2-2x-3






My attempt solution is differ x^2 - 2x + 4 = 2x - 2 and basically the top x-1 can be manipulate by multiplying 2 this would give me the answer as 2 ln|x^2-2x+4| + c I am not sure I am right can somebody help me out? thank you!

 

[MostlyHarmless]

You are almost correct. This is where you went wrong, you should recognize this as a simple u-substitution. So you would set u=x^2-2x-3 and du=2x-2 dx. And you were correct in recognizing that it only differs the numerator by a constant, but what you would want to do is now divide du=2x-2 by 2. This gives (1/2)du=x-1dx. Now you can make the substitution. And you get $(1/2)[integral]du/u= (1/2)ln(u)+c$ then just plug back in for u and there's your answer.

Just try to physically write down the u substitution and everything that follows and you won't lose coefficients like that.

 

[luigihs]

thanks ! sorry and for this one ∫x-3 / x^2+2x-3 .. i think can't do the same as the previous one because would be bottom 2x + 2 but in the top is x - 3 so is a bit impossible to do the previous method. but can i do by partial fractions (x+3) (x-1)

 

[Ray Vickson]

thanks ! sorry and for this one ∫x-3 / x^2+2x-3 .. i think can't do the same as the previous one because would be bottom 2x + 2 but in the top is x - 3 so is a bit impossible to do the previous method. but can i do by partial fractions (x+3) (x-1)

You should have no trouble computing
\int \left( x - \frac{3}{x^2} + 2x - 3 \right) \, dx
which is what you wrote. If you meant something else, you must use parentheses.

 

[MostlyHarmless]

thanks ! sorry and for this one ∫x-3 / x^2+2x-3 .. i think can't do the same as the previous one because would be bottom 2x + 2 but in the top is x - 3 so is a bit impossible to do the previous method. but can i do by partial fractions (x+3) (x-1)

This one is the two parts, the first part will be the exact same as the previous integral. The next part will be just as easy. How can you break this into two integrals one of which is exactly the same as the previous problem?