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Partial fractions in integration (view source)
Revision as of 14:30, December 30, 2008
, 14:30, December 30, 2008→A Repeated 1st-Degree Denominator: reworded and clarified a few sentences
<br /><math>u = ax+b</math>
<br /><math>u = ax+b</math>
<br /><math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-k} + C = {-1 \over kau^{k-1}} + C = {-1 \over {ka (ax+b)^{k-1}}} + C</math>
<br /><math>= \int \frac{1}{u^k} \frac{du}{a} = \frac{1}{a} \int u^{-k} du = \frac{1}{a} \cdot \frac{u^{k-1}}{-k} + C = {-1 \over kau^{k-1}} + C = {-1 \over {ka (ax+b)^{k-1}}} + C</math>
<br />Note that the above formula '''only''' works if <math>k>1</math>. It does not matter if <math>k</math> is a fraction or even irrational, just as long as it is '''greater than 1'''.
<br />This means that integrals such as
<br />This means that integrals such as
<br /><math>\int {5 \over {(3x+17)^{12}}}dx</math>
<br /><math>\int {5 \over {(3x+17)^{12}}}dx</math>