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===Partial Fractions===
 
===Partial Fractions===
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'''Integration by partial fractions''' is a [[Techniques of integration|technique]] to facilitate the integration of a rational expression by partial fraction decomposition.
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Given an integral
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:<big><math>\int\frac{3x+11}{x^2-x-6}dx</math></big>
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The first step is to factor the denominator as much as possible and get the form of the partial fraction decomposition.  Doing this gives,
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:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}</math></big>
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This allows us to split the fraction in to sums by cross multiplying the denominators,
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:<big><math>\frac{3x+11}{(x-3)(x+2)}=\frac{A(x+2)+B(x-3)}{(x-3)(x+2)}</math></big>
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Therefore we can restate the problem,
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:<big><math>3x+11=A(x+2)+B(x-3)</math></big>
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Now we can solve for A and B by subsituting x with a value that allows the term to go to 0.  For example,
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We let :<big><math>x=-2</math></big>,
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:<big><math>3(-2)+11=A(-2+2)+B(-2-3)</math></big>
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:<big><math>5=A(0)+B(5)</math></big>
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:<big><math>B=-1</math></big>
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We let :<big><math>x=3</math></big>,
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:<big><math>3(3)+11=A(3+2)+B(3-3)</math></big>
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:<big><math>20=A(5)+B(0)</math></big>
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:<big><math>A=4</math></big>
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We then plug in the values of A and B and get,
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:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>
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Now we can solve the integral.
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:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}-\frac{1}{x+2}dx</math></big>
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:<big><math>\int\frac{3x+11}{x^2-x-6}dx=\int\frac{4}{x-3}dx-\int\frac{1}{x+2}dx</math></big>
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:<big><math>\int\frac{3x+11}{x^2-x-6}dx=4ln|x-3|-ln|x+2|+c</math></big>
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===Algebraic Substitution===
 
===Algebraic Substitution===
 
===Trigonometric Substitution===
 
===Trigonometric Substitution===
nsJudgesRO, nsJudgesRW, nsJudges_talkRO, nsJudges_talkRW
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