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Calculating the integral of a function of the form $\\frac{c}{(x+a)(x+b)}$ using partial fractions.

", "licence": "None specified"}, "statement": "

Calculate the integral

\n

\\[  \\simplify{int({c}/((x+{a})(x+{b})),x)} .\\]

", "advice": "

In order to integrate the function \\[ \\simplify{{c}/((x+{a})(x+{b}))}, \\] we want to rewrite it in terms of its partial fractions.

\n

To do this, we want to set the function equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{x+{b}}$. Since these are both distinct linear factors, this tells us that the numerators will be constants, which we will call $A$ and $B$:

\n

\\[ \\simplify{{c}/((x+{a})(x+{b}))} = \\simplify{A/(x+{a}) + B/(x+{b})}.\\]

\n

To find the values of $A$ and $B$, we want to multiply this equation by the denominator of the left-hand side. This gives

\n

\\[ \\simplify{{c}=A(x+{b})+B(x+{a})}.\\]

\n

\n

To find $A$, we can eliminate $B$ by setting $\\simplify{x={-a}}$:

\n

\\[ \\simplify{{c}=A{b-a}} \\implies \\simplify[fractionNumbers]{A={c/(b-a)}}.\\]

\n

Similarly, to find B, we can eliminate $A$ by setting $\\simplify{x={-b}}$:

\n

\\[ \\simplify{{c}=B{a-b}} \\implies \\simplify[fractionNumbers]{B={c/(a-b)}}.\\]

\n

Therefore, 

\n

{check1}

\n

and

\n

{check2}

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{Asol}/(x+{a})+{Bsol}/(x+{b})},\\\\]

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\\\\[ \\\\simplify{{c}/((x+{a})(x+{b}))} = \\\\simplify[all,fractionNumbers]{{c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b}))},\\\\]

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\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,=  \\\\simplify[all,fractionNumbers]{int({Asol}/(x+{a})+{Bsol}/(x+{b}),x)}\\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)}   \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[all,fractionNumbers]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

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\\\\[ \\\\begin{split} \\\\simplify{int({c}/((x+{a})(x+{b})),x)} &\\\\,= \\\\simplify[all,fractionNumbers]{int({c}/(({b-a})(x+{a}))+{c}/(({a-b})(x+{b})),x)} \\\\\\\\\\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} int(1/(x+{a}),x)+{Bsol} int(1/(x+{b}),x)}   \\\\\\\\ \\\\\\\\ &\\\\,=\\\\simplify[basic,fractionNumbers,zeroFactor,noLeadingMinus]{{Asol} ln (abs(x+{a}))+{Bsol} ln (abs(x+{b})) + C}. \\\\end{split}\\\\]

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