Calculus BC ยท Unit 6
Match the shape of the integrand to a method, then finish the integral.
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Look at the shape before you compute
A basic antiderivative needs no extra method. A composite with its derivative sitting beside it is substitution. A product of unrelated types is often parts.
A rational function has its own path
Divide if the top is too tall. Factor distinct linear factors for partial fractions. Complete the square if the quadratic has no real roots.
An infinite limit is a separate question
First decide whether the integral converges. Only a convergent integral has a value.
Name the method, then do the short version
On a multiple-choice item, a wrong method often produces a familiar wrong answer. Check that the derivative of your result returns the integrand.
Name the method in the same breath as the answer.
This is a sum of basic powers. No substitution or parts is needed.
\(x^4+2x+C\)
Answer: \(x^4+2x+C\)
\(du=2x\,dx\) is already the extra factor.
The integral becomes \(\int e^u\,du\).
Answer: A. substitution with \(u=x^2\)
\(u=x^2\)
\(du=2x\,dx\)
\(e^{x^2}+C\)
Answer: \(e^{x^2}+C\)
\(x\) and \(e^x\) are different types. Their derivatives are not hiding in each other.
Parts with \(u=x\) leaves \(\int e^x\,dx\).
Answer: A. parts with \(u=x\)
This is a constant times \(1/u\) with \(u=x+1\).
\(6\ln|x+1|+C\)
Answer: \(6\ln|x+1|+C\)
\(p=4>1\)
The power tail converges.
Its value is \(\left[-\dfrac{1}{3x^3}\right]_1^{\infty}=\dfrac{1}{3}\).
Answer: converges to \(\dfrac{1}{3}\)
Two plausible methods. Only one keeps the algebra short.
The degrees match, so rewrite before integrating.
\(\dfrac{x}{x+1}=1-\dfrac{1}{x+1}\)
\(x-\ln|x+1|+C\)
Answer: \(x-\ln|x+1|+C\)
Two distinct linear factors: partial fractions.
\(4=A(x+1)+B(x-3)\)
At \(x=3\), \(A=1\). At \(x=-1\), \(B=-1\).
\(\ln|x-3|-\ln|x+1|+C\)
Answer: \(\ln\left|\dfrac{x-3}{x+1}\right|+C\)
Substitution, not parts. \(u=x^2\) and \(du=2x\,dx\).
\(-\dfrac{1}{2}\cos(x^2)+C\)
Answer: \(-\dfrac{1}{2}\cos(x^2)+C\)
The upper limit is infinite, so write a limit.
\(\lim_{b\to\infty}\left[-e^{-4x}\right]_0^b\)
\(0-(-1)\)
\(1\)
Answer: \(1\)
The numerator is the derivative of the inside of the denominator, up to the factor already present.
\(u=x^2+9\)
\(du=2x\,dx\)
\(\ln(x^2+9)+C\)
Answer: \(\ln(x^2+9)+C\)
\(\cos(4x)\) is a basic chain-rule antiderivative.
\(x/(x^2+1)\) is a logarithm substitution.
\(1/(x^2+1)\) is \(\arctan x\).
Only \(x\cos x\) is a product of two unrelated types.
Answer: A. \(\int x\cos x\,dx\)
The first step changes the later method.
Divide term by term. This is not partial fractions yet.
\(x+1+\dfrac{1}{x}\)
\(\dfrac{x^2}{2}+x+\ln x+C\)
Answer: \(\dfrac{x^2}{2}+x+\ln x+C\)
No real factors: \((x+3)^2+4\).
\(\int \dfrac{1}{u^2+4}\,du\)
\(\dfrac{1}{2}\arctan\dfrac{x+3}{2}+C\)
Answer: \(\dfrac{1}{2}\arctan\dfrac{x+3}{2}+C\)
Parts: \(u=x\), \(dv=e^{-x}\,dx\), \(v=-e^{-x}\).
\(\lim_{b\to\infty}\left[-xe^{-x}\right]_1^b-\lim_{b\to\infty}\int_1^b -e^{-x}\,dx\)
The boundary term at infinity is 0, and at 1 it is \(-e^{-1}\).
\(0-\left(-\dfrac{1}{e}\right)+\lim_{b\to\infty}\left[-e^{-x}\right]_1^b\)
\(\dfrac{1}{e}+\left(0-\left(-\dfrac{1}{e}\right)\right)\)
\(\dfrac{2}{e}\)
Answer: \(\dfrac{2}{e}\)
\(5x-1=A(x+4)+B(x-1)\)
At \(x=1\): \(4=5A\), so \(A=4/5\).
At \(x=-4\): \(-21=-5B\), so \(B=21/5\).
\(\dfrac{4}{5}\ln|x-1|+\dfrac{21}{5}\ln|x+4|+C\)
Answer: \(\dfrac{4}{5}\ln|x-1|+\dfrac{21}{5}\ln|x+4|+C\)
The integrand is continuous on \([0,1)\) and unbounded at \(x=1\), but the substitution may remove that.
\(u=1-x^2\)
\(du=-2x\,dx\)
When \(x=0\), \(u=1\). When \(x=1\), \(u=0\).
\(\dfrac{1}{2}\int_0^1 u^{-1/2}\,du\)
\(\dfrac{1}{2}\left[2u^{1/2}\right]_0^1\)
\(1\)
The improper integral converges, and substitution evaluates it.
Answer: \(1\)
Divide first: \(\dfrac{x^2}{x^2+1}=1-\dfrac{1}{x^2+1}\).
\(\int 1\,dx=x\)
\(\int \dfrac{1}{x^2+1}\,dx=\arctan x\)
\(x-\arctan x+C\)
Answer: A. \(x-\arctan x+C\)
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