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Last edited by LuisRodg (2008-02-18 13:11:46)
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Use partial fractions:
Now use a linear u-substitution on the first fraction and the other two fractions are easy.
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We havent got to partial fractions. Thats next section. We are supposed to solve it without partial fractions?
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That's not really partial fractions though. You split a fraction into two partial fractions if the denominator has a higher order than the numerator.
What's happening here is you're just factoring out whole numbers and leaving a constant as the numerator of your fraction (thus making it integrable).
Why did the vector cross the road?
It wanted to be normal.
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