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Calculating the integral of a function of the form $\\frac{cx+d}{(x+a)^2}$ using partial fractions.

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Calculate the integral

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\\[ \\simplify{int(({c}x+{d})/(x+{a})^2,x)} .\\]

", "advice": "

In order to integrate the function \\[ \\simplify{({c}x+{d})/(x+{a})^2}, \\] we want to rewrite it in terms of its partial fractions.

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Since we have a repeated linear factor we want to set the function equal to the sum of 2 fractions with denominators $\\simplify{x+{a}}$ and $\\simplify{(x+{a})^2}$. The numerators will be constants, which we will call $A$ and $B$:

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\\[ \\simplify{({c}x+{d})/((x+{a})^2)} = \\simplify{A/(x+{a}) + B/(x+{a})^2}\\]

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

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

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By comparing the coefficients of the $x$-terms and the constant terms we can form a pair of simultaneous equations to find $A$ and $B$:

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\\[ \\begin{split} \\simplify{{c}x+{d}} &\\,= \\simplify{A(x+{a})+B} \\\\ &\\,= \\simplify{A*x+{a}A+B}. \\end{split} \\]

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\\[ \\begin{split}&(x):\\quad \\var{c} &\\,= \\simplify{A} \\\\ &(c):\\quad \\var{d} &\\,= \\simplify{{a}A+B} .\\end{split} \\]

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Hence,

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\\[A=\\var{c},\\,B=\\simplify{{d-a*c}}, \\]

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and

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\\[ \\simplify{({c}x+{d})/((x+{a})^2)} = \\simplify{{c}/(x+{a}) + {d-a*c}/(x+{a})^2}.\\]

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Therefore,

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\\[ \\begin{split} \\simplify{int(({c}x+{d})/((x+{a})^2),x)} &\\,= \\simplify{int({Asol}/(x+{a}) + {Bsol}/(x+{a})^2,x)} \\\\ &\\,=\\simplify[basic,fractionNumbers,noLeadingMinus]{{Asol} int(1/(x+{a}),x)+{Bsol} int((x+{a})^-2,x)} \\\\\\\\ &\\,=\\simplify[all,fractionNumbers]{{Asol} ln(abs(x+{a}))-{Bsol}/(x+{a})+C}. \\end{split}\\]

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