fincle1
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Can someone explain how i should integrate this:
\displaystyle \int_2^3 [\frac{x}{5-x^2}] dx.
The way that I tried gives: \left[-\frac{1}{2}ln(5-x^2)]\right_2^3
which, when I insert the values to get:
[-\frac{1}{2}ln(5-3^2)]-[-\frac{1}{2}ln(5-2^2)]
I end up with my calculator telling me there's an error.
I'm guessing this error is because: \[-\frac{1}{2}ln(5-x^2) isn't possible.
Can someone explain where/if i'm going wrong and how I should approach this?

Thanks.
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the bear
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the problem is that the underneath will reach zero somewhere between the limits 2 and 3
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Hasufel
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Because the denominator is infinite between llimits, as thebear indicated, you have to use "improper integrals" and limits. Try expressing the integrand as:

\displaystyle \frac{x}{(\sqrt{5}+x)(\sqrt{5}-x)}

using partial fractions, you can then have 3 integrals - the 2nd 2 improper - the first one having limits between 2 and 3, the 2nd between 2 and root(5)-a small arbitrary number, epsilon, and 3rd one having limits between root(5)+a small arbitrary number epsilon.

then, evaluate the first by taking:

\displaystyle \int_{2}^{3} \frac{A}{(\sqrt{5}+x)}

and the 2nd one as:


\displaystyle lim_{ \epsilon ->0} \int_{2}^{\sqrt{5}-\epsilon} \frac{B}{(\sqrt{5}-x)}+ lim_{ \epsilon ->0} \int_{\sqrt{5}+\epsilon}^{3} \frac{B}{(\sqrt{5}-x)}

(admiteddly, evaluating the integrand without limits similar to the way you have is easier, then taking limits as i have at the points of infinity is speedier)
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noodledoodle
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Plot F(x) into your GC and get the graph. You will see that x=5^1/2 is a vertical asymptote as y tends to infinity
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TenOfThem
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(Original post by fincle1)
Can someone explain how i should integrate this:
\displaystyle \int_2^3 [\frac{x}{5-x^2}] dx.
The way that I tried gives: \left[-\frac{1}{2}ln(5-x^2)]\right_2^3
which, when I insert the values to get:
[-\frac{1}{2}ln(5-3^2)]-[-\frac{1}{2}ln(5-2^2)]
I end up with my calculator telling me there's an error.
I'm guessing this error is because: \[-\frac{1}{2}ln(5-x^2) isn't possible.
Can someone explain where/if i'm going wrong and how I should approach this?

Thanks.
In addition to the other answers referring to the asymptote

When we integrate a function that leads to ln gives ln|f(x)| so -\frac{1}{2}\ln |5-x^2|
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