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Practice question simplifying fractions for integration by using partial fractions.
", "licence": "All rights reserved"}, "statement": "Simplify the following expression using partial fractions and find the general solution.
\n\n$\\frac{\\var{a}x\\var{b}}{(x-\\var{c})(x+\\var{d})}$
", "advice": "The following must be decomposed by partial fractions to be integrated
\n$\\frac{\\var{a}*x\\var{b}}{(x-\\var{c})(x+\\var{d})}$
\nThe denominator contains two linear factors, $(x-\\var{c})$ and $(x+\\var{d})$.
\nThus, the decomposition of the original expression is as follows
\n$\\frac{\\var{a}*x\\var{b}}{(x-\\var{c})(x+\\var{d})} = \\frac{A}{x-\\var{c}} + \\frac{B}{x+\\var{d}}$
\n\nAs both factors are linear, the cover-up rule can be used:
\n$A\\lim=\\frac{\\var{a}*\\var{c}\\var{b}}{(\\var{c}+\\var{d})}$ = \\[\\var{((a*c)+b)/((c)+d)}\\]
\n$B\\lim=\\frac{\\var{a}*\\var{d}\\var{b}}{(\\var{d}-\\var{c})}$ = \\[\\var{((a*(-d))+b)/((-d)-c)}\\]
\nSub back into the partial fractions
\n\\[\\frac{\\var{((a*c)+b)/((c)+d)}}{x-\\var{c}} + \\frac{\\var{((a*(-d))+b)/((-d)-c)}}{x+\\var{d}}\\]
\n\nThe expressions can now be integrated
\n\\[\\int{\\frac{\\var{((a*c)+b)/((c)+d)}}{x-\\var{c}} + \\frac{\\var{((a*(-d))+b)/((-d)-c)}}{x+\\var{d}}}.dx\\]
\nIntegration by parts
\n\\[\\int{\\frac{\\var{((a*c)+b)/((c)+d)}}{x-\\var{c}}}.dx + \\int{\\frac{\\var{((a*(-d))+b)/((-d)-c)}}{x+\\var{d}}}.dx\\]
\nMove the constants outside
\n\\[\\var{((a*c)+b)/((c)+d)}\\int{\\frac{1}{x-\\var{c}}}.dx + \\var{((a*(-d))+b)/((-d)-c)}\\int{\\frac{1}{x+\\var{d}}}.dx\\]
\nSubstitution
\nlet $u=x-\\var{c}$ therefore $\\frac{du}{dx}=1 \\therefore dx=du$
\nlet $v=x+\\var{d}$ therefore $\\frac{dv}{dx}=1 \\therefore dx=dv$
\nSub these substitutions in to the integrals
\n\\[\\var{((a*c)+b)/((c)+d)}\\int{\\frac{1}{u}}.du + \\var{((a*(-d))+b)/((-d)-c)}\\int{\\frac{1}{v}}.dv\\]
\nIntegration then yields
\n\\[ \\var{((a*c)+b)/((c)+d)} ln(u) + \\var{((a*(-d))+b)/((-d)-c)}ln(v) +C \\]
\nSub back in the original denominators for the general solution
\n\\[ \\var{((a*c)+b)/((c)+d)} ln(x-\\var{c}) + \\var{((a*(-d))+b)/((-d)-c)}ln(x+\\var{d}) +C \\]
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\n\n$\\frac{\\var{a}x\\var{b}}{(x-\\var{c})(x+\\var{d})}$=[[0]]
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\n\n$A =$ [[0]] and $B =$[[1]]
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\n$\\frac{\\var{a}x\\var{b}}{(x-\\var{c})(x+\\var{d})}$=[[0]]
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\n\n\\[\\int{\\frac{\\var{((a*c)+b)/((c)+d)}}{x-\\var{c}}}.dx + \\int{\\frac{\\var{((a*(-d))+b)/((-d)-c)}}{x+\\var{d}}}.dx\\]
\n\nUse C as the constant of integration.
\n$y=$[[0]]
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