jamesbob said:
Cheers!
When doing the polynomial division i got
(x^2 - 1) + \frac{1}{x^2 + 1}
Is this correct. If so, what's the rule for applying partial fractions to an irreducible again? i.e. we have
x^2 + 1 so x^2 = -1
No, you cannot use
partial fraction on that.
Your denominator is (x
2 + 1), it cannot be factored any further.
For short, if your denominator is a product of
some polynomials of degree 2 or 1, you can use partial fraction. Note that every polynimial can be factored to be the product of some polynomials of degree less than or equal to 2.
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\int\frac{1}{1 + x^2} = tan^{-1}x
Correct, \int\frac{dx}{1 + x^2} = \arctan x + C
Don't forget
dx, and +C.
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Is it possible to integrate the function? If not, how do you plug the x(-3) in?
Yes, it is. It's possible to integrate that. Now the degree of the numerator is
equal to the degree of the denominator, what should you do first? (Hint: Polynomial division).
From there, I think you can solve the problem.
I'll give you an example:
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Example 1:
\int \frac{dx}{x ^ 2 + 2x + 3} = \int \frac{dx}{(x + 1) ^ 2 + 2}
Use u-substitution:
u = x + 1 => du = dx
We have:
\int \frac{dx}{(x + 1) ^ 2 + 2} = \int \frac{du}{u ^ 2 + (\sqrt{2}) ^ 2} = \frac{1}{\sqrt{2}} \arctan \left( \frac{u}{\sqrt{2}} \right) + C. Change u back to x, we have:
\int \frac{dx}{x ^ 2 + 2x + 3} = \frac{1}{\sqrt{2}} \arctan \left( \frac{x + 1}{\sqrt{2}} \right) + C
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Example 2:
\int \frac{x}{x ^ 2 + x + 3} dx
Take the derivative of the denominator with respect to x, we have:
(x
2 + x + 3)' = 2x + 1.
Now:
\int \frac{x}{x ^ 2 + x + 3} dx = \frac{1}{2} \int \frac{2x}{x ^ 2 + x + 3} dx = \frac{1}{2} \int \frac{2x + 1 - 1}{x ^ 2 + x + 3} dx
= \frac{1}{2} \int \frac{2x + 1}{x ^ 2 + x + 3} dx - \frac{1}{2} \int \frac{1}{x ^ 2 + x + 3} dx
To integrate this:
\int \frac{2x + 1}{x ^ 2 + x + 3} dx
Let u = x
2 + x + 3, that means du = 2x + 1 dx, so we have:
\int \frac{du}{u} = \ln |u| + C = \ln|x ^ 2 + x + 3| + C = \ln(x ^ 2 + x + 3) + C, we can drop the absolute value because: x
2 + x + 3 > 0, for all x.
To integrate:
\int \frac{1}{x ^ 2 + x + 3} dx = \int \frac{dx}{\left(x + \frac{1}{2} \right) ^ 2 + \frac{11}{4}}. Now do as example 1, let u = x + 1 / 2, we have:
\int \frac{1}{x ^ 2 + x + 3} dx = \frac{2}{\sqrt{11}} \arctan \left( 2 \times \frac{u}{\sqrt{11}} \right) + C = \frac{2}{\sqrt{11}} \arctan \left( 2 \times \frac{x + \frac{1}{2}}{\sqrt{11}} \right)
= \frac{2}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C.
So we have:
\int \frac{x}{x ^ 2 + x + 3} dx = \frac{1}{2} \ln(x ^ 2 + x + 3) - \frac{1}{2} \times \frac{2}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C
= \frac{1}{2} \ln(x ^ 2 + x + 3) - \frac{1}{\sqrt{11}} \arctan \left( \frac{2x + 1}{\sqrt{11}} \right) + C
Now let's see if you can work out that integral. :)
