Calculus BC · Unit 6
Divide first when you need to, split distinct linear factors, and finish a solid quadratic with an arctangent.
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Divide when the top is at least as tall as the bottom
If the numerator’s degree is greater than or equal to the denominator’s degree, divide first.
Write the integrand as a polynomial plus a proper fraction.
Distinct linear factors split
\(\dfrac{px+q}{(x-a)(x-b)}=\dfrac{A}{x-a}+\dfrac{B}{x-b}\)
Cover one factor, or substitute the root, to find each constant.
Complete the square for a quadratic that does not factor
\(x^2+bx+c=(x+\tfrac{b}{2})^2+\text{a positive number}\) leads to an arctangent, or to a logarithm if the numerator matches the derivative.
Split a numerator that is almost a derivative
Write the numerator as a multiple of the derivative of the denominator, plus a constant leftover. Integrate the two pieces separately.
One division, or two linear factors with easy constants.
The top has higher degree, so divide.
\((x+1)(x+2)=x^2+3x+2\)
\(x^2+3x+5=(x+1)(x+2)+3\)
\(\dfrac{x^2+3x+5}{x+1}=x+2+\dfrac{3}{x+1}\)
\(\dfrac{x^2}{2}+2x+3\ln|x+1|+C\)
Answer: \(\dfrac{x^2}{2}+2x+3\ln|x+1|+C\)
\(\dfrac{5}{(x-2)(x+3)}=\dfrac{A}{x-2}+\dfrac{B}{x+3}\)
\(5=A(x+3)+B(x-2)\)
At \(x=2\): \(5=5A\), so \(A=1\).
At \(x=-3\): \(5=-5B\), so \(B=-1\).
\(\ln|x-2|-\ln|x+3|+C\)
Answer: \(\ln\left|\dfrac{x-2}{x+3}\right|+C\)
\(\dfrac{1}{x(x+1)}=\dfrac{A}{x}+\dfrac{B}{x+1}\)
\(1=A(x+1)+Bx\)
At \(x=0\), \(A=1\). At \(x=-1\), \(B=-1\).
\(\ln|x|-\ln|x+1|+C\)
Answer: \(\ln\left|\dfrac{x}{x+1}\right|+C\)
\(x^2+4x+5=(x+2)^2+1\)
\(u=x+2\)
\(\int \dfrac{1}{u^2+1}\,du\)
\(\arctan(x+2)+C\)
Answer: \(\arctan(x+2)+C\)
\(x+2=(x+4)-2\)
\(\dfrac{x+2}{x+4}=1-\dfrac{2}{x+4}\)
\(x-2\ln|x+4|+C\)
Answer: \(x-2\ln|x+4|+C\)
\(2x+1=A(x+2)+B(x-1)\)
At \(x=1\): \(3=3A\), so \(A=1\).
At \(x=-2\): \(-3=-3B\), so \(B=1\).
\(\dfrac{1}{x-1}+\dfrac{1}{x+2}\)
Answer: \(\dfrac{1}{x-1}+\dfrac{1}{x+2}\)
Integrate the split, or scale the arctangent.
From the cover-up step, the split is \(\dfrac{1}{x-1}+\dfrac{1}{x+2}\).
\(\int \dfrac{1}{x-1}\,dx=\ln|x-1|\)
\(\int \dfrac{1}{x+2}\,dx=\ln|x+2|\)
\(\ln|x-1|+\ln|x+2|+C\)
Answer: \(\ln|(x-1)(x+2)|+C\)
\(x^2+2x+5=(x+1)^2+4\)
\(u=x+1\)
\(\int \dfrac{1}{u^2+4}\,du\)
\(\dfrac{1}{2}\arctan\dfrac{u}{2}+C\)
\(\dfrac{1}{2}\arctan\dfrac{x+1}{2}+C\)
Answer: \(\dfrac{1}{2}\arctan\dfrac{x+1}{2}+C\)
The numerator is half the derivative of the denominator.
Derivative of the bottom: \(2x+2=2(x+1)\).
\(\dfrac{1}{2}\int \dfrac{2x+2}{x^2+2x+5}\,dx\)
\(\dfrac{1}{2}\ln(x^2+2x+5)+C\)
Answer: \(\dfrac{1}{2}\ln(x^2+2x+5)+C\)
\(x^2=(x+2)(x-2)+4\)
\(\dfrac{x^2}{x+2}=x-2+\dfrac{4}{x+2}\)
\(\dfrac{x^2}{2}-2x+4\ln|x+2|+C\)
Answer: \(\dfrac{x^2}{2}-2x+4\ln|x+2|+C\)
\(3x=A(x+1)+B(x-2)\)
At \(x=2\): \(6=3A\), so \(A=2\).
At \(x=-1\): \(-3=-3B\), so \(B=1\).
\(2\ln|x-2|+\ln|x+1|+C\)
Answer: \(\ln\left|(x-2)^2(x+1)\right|+C\)
\(\dfrac{1}{x(x+2)}=\dfrac{1/2}{x}-\dfrac{1/2}{x+2}\)
Check: \(\dfrac{1}{2}\cdot\dfrac{(x+2)-x}{x(x+2)}=\dfrac{1}{x(x+2)}\).
\(\dfrac{1}{2}\left[\ln x-\ln(x+2)\right]_2^4\)
At 4: \(\ln\dfrac{4}{6}=\ln\dfrac{2}{3}\).
At 2: \(\ln\dfrac{2}{4}=\ln\dfrac{1}{2}\).
\(\dfrac{1}{2}\left(\ln\dfrac{2}{3}-\ln\dfrac{1}{2}\right)\)
\(\dfrac{1}{2}\ln\dfrac{4}{3}\)
Answer: \(\dfrac{1}{2}\ln\dfrac{4}{3}\)
A leftover constant after the logarithm piece, or a denominator with no real roots.
Derivative of the bottom: \(2x+2\).
\(2x+3=(2x+2)+1\)
\(\int \dfrac{2x+2}{x^2+2x+5}\,dx=\ln(x^2+2x+5)\)
The leftover is \(\int \dfrac{1}{(x+1)^2+4}\,dx\).
\(\dfrac{1}{2}\arctan\dfrac{x+1}{2}\)
\(\ln(x^2+2x+5)+\dfrac{1}{2}\arctan\dfrac{x+1}{2}+C\)
Answer: \(\ln(x^2+2x+5)+\dfrac{1}{2}\arctan\dfrac{x+1}{2}+C\)
\(x^2-x-2=(x-2)(x+1)\)
\(4x-2=A(x+1)+B(x-2)\)
At \(x=2\): \(6=3A\), so \(A=2\).
At \(x=-1\): \(-6=-3B\), so \(B=2\).
\(2\ln|x-2|+2\ln|x+1|+C\)
Answer: \(2\ln|(x-2)(x+1)|+C\)
\(x^2+1=(x^2+4)-3\)
\(\dfrac{x^2+1}{x^2+4}=1-\dfrac{3}{x^2+4}\)
\(x-\dfrac{3}{2}\arctan\dfrac{x}{2}+C\)
Answer: \(x-\dfrac{3}{2}\arctan\dfrac{x}{2}+C\)
\(x^2+2x+2=(x+1)^2+1\)
\(\left[\arctan(x+1)\right]_0^1\)
\(\arctan 2-\arctan 1\)
\(\arctan 2-\dfrac{\pi}{4}\)
Answer: \(\arctan 2-\dfrac{\pi}{4}\)
The denominator \((x+2)^2+1\) has no real linear factors.
Its derivative is \(2x+4\), while the numerator is \(x+1=(x+2)-1\).
Split into a logarithm piece and an arctangent piece.
Answer: A. complete the square, then split the numerator
\(x^2+4x+5=(x+2)^2+1\)
\(x+1=(x+2)-1\)
Logarithm piece: \(\dfrac{1}{2}\ln(x^2+4x+5)\).
Leftover: \(-\int \dfrac{1}{(x+2)^2+1}\,dx\).
\(\dfrac{1}{2}\ln(x^2+4x+5)-\arctan(x+2)+C\)
Answer: \(\dfrac{1}{2}\ln(x^2+4x+5)-\arctan(x+2)+C\)
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