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This exercise is best solved by using substitution.

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Note that the numerator $\\simplify[std]{{2 * a} * x + {b}}$ of \\[\\simplify[std]{({2 * a} * x + {b}) / ({a} * x ^ 2 + {b} * x + {c})}\\] is the derivative of the denominator $\\simplify[std]{{a} * x ^ 2 + {b} * x + {c}}$

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So if you use as your substitution $u=\\simplify[std]{{a} * (x ^ 2) + ({b} * x) + {c}}$ you then have $\\simplify[std]{ du = ({2 * a} * x + {b}) * dx}$

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Hence we can replace $\\simplify[std]{ ({2 * a} * x + {b}) * dx}$ by $du$

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Hence the integral becomes:

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\\[\\begin{eqnarray*} I&=&\\int\\;\\frac{du}{u}\\\\ &=&\\ln(|u|)+C\\\\ &=& \\simplify[std]{ln(abs({a} * (x ^ 2) + ({b} * x) + {c}))+C} \\end{eqnarray*}\\]

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A Useful Result
This example can be generalised.

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Suppose \\[I = \\int\\; \\frac{f'(x)}{f(x)}\\;dx\\]

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The using the substitution $u=f(x)$ we find that $du=f'(x)\\;dx$ and so using the same method as above:
\\[I = \\int \\frac{du}{u} = \\ln(|u|)+ C = \\ln(|f(x)|)+C\\]

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Find the following integral.

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You must input the constant of integration as $C$.

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Input all numbers as integers or fractions.

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You can click on Show steps to get a hint. You will lose 1 mark if you do so.

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Note that $\\displaystyle \\int \\frac{1}{x}\\;dx=\\ln(|x|)+C$ and you must include the absolute value in the argument of $\\ln$. You input $|x|$ as abs(x).

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Find $\\displaystyle I=\\int \\frac{2 a x + b} {a x ^ 2 + b x + c}\\;dx$ by substitution or otherwise.

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