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Line 11: − <br /><math>\int \frac{1}{ax + b}dx</math>+ − <br />Substitute <math>u = ax + b</math>+ − <br /><math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>+ − <br /><math>\int \frac{8}{3x+13}dx</math>+ − <br />The steps can be skipped by using the general formula above get:+ − <br /><math>= \frac{8}{3} \ln{|3x+13|}+C</math>+ − + − <br /><math>\int \frac{1}{(ax+b)^k}dx</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 />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 /><math>\int {5 \over {(3x+17)^{12}}}dx</math>+ − <br />are now very easy:+ − <br /><math>= {-5 \over {(3)(12)(3x+17)^{11}}}+C = {-5 \over {36(3x+17)^{11}}}+C</math>+ Line 52:
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Partial fractions in integration (view source)
Revision as of 18:11, December 19, 2018
, 18:11, December 19, 2018Remove extra whitespace, wikify and indent math expressions
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'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the integration of a rational expression by partial fraction decomposition.
'''Integration by partial fractions''' is a [[Techniques of integration|technique]] in [[Calculus]] to facilitate the [[integration]] of a rational expression by partial fraction decomposition. Given an integral
:<math>\int \frac {f(x)}{g(x)}dx</math>
where <math>f(x)</math> and <math>g(x)</math> are both [[polynomial]]s, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
Given an integral
<math>\int \frac {f(x)}{g(x)}dx</math>
where <math>f(x)</math> and <math>g(x)</math> are both polynomials, integration by partial fractions shows how to separate the problem into multiple integrals before integrating.
==Integration by Partial Fractions==
==Integration by Partial Fractions==
=== A 1st-Degree Denominator ===
=== A 1st-Degree Denominator ===
Given:
Given:
:<math>\int \frac{1}{ax + b}dx</math>
Substitute <math>u = ax + b</math>
:<math>= \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \int \frac{du}{u} = \frac{1}{a} \ln{|u|} + C = \frac{1}{a} \ln{|ax + b|} + C</math>
This means that if given an integral such as:
This means that if given an integral such as:
:<math>\int \frac{8}{3x+13}dx</math>
The steps can be skipped by using the general formula above get:
:<math>= \frac{8}{3} \ln{|3x+13|}+C</math>
=== A Repeated 1st-Degree Denominator ===
=== A Repeated 1st-Degree Denominator ===
The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.
The formula for integrals where a first degree polynomial denominator is raised to a power greater than one is much different than the formula above.
Given
Given:
:<math>\int \frac{1}{(ax+b)^k}dx</math>
:<math>u = ax+b</math>
:<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-1)} + C = {-1 \over (k-1)au^{k-1}} + C = {-1 \over {(k-1)a (ax+b)^{k-1}}} + C</math>
Note that the above formula '''only''' works if <math>k \neq 1</math>.
<br />This means that integrals such as
<br />This means that integrals such as
:<math>\int {5 \over {(3x+17)^{12}}}dx</math>
are now very easy:
:<math>= {-5 \over {(3)(11)(3x+17)^{11}}}+C = {-5 \over {33(3x+17)^{11}}}+C</math>
== A 2nd-Degree Denominator ==
== A 2nd-Degree Denominator ==
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
Therefore the problem can be restated,
Therefore, the problem can be restated,
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
==See Also==
==See also==
*[[Calculus]]
*[[Calculus]]
*[[Integral]]
*[[Integral]]
*[[Methods of integration]]
*[[Methods of integration]]
[[Category:calculus]]
[[Category:Calculus]]
[[Category:integration]]
[[Category:Integration]]