Integration by substitution, by parts and with partial fractions flashcards

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Integration by substitution
Rewriting an integral in a new variable u = g(x), replacing every x and the dx, so that it becomes a standard integral in u. It reverses the chain rule.
Integration by parts
A method for integrating a product: ∫u (dv/dx) dx = uv − ∫v (du/dx) dx. It reverses the product rule.
Partial fractions (for integration)
Writing a rational function as a sum of simpler fractions with linear (or repeated linear) denominators, each of which can be integrated.
Improper algebraic fraction
A fraction whose numerator has degree greater than or equal to that of the denominator; divide first to get a polynomial part plus a proper fraction.
Changing the limits
Replacing the x-limits of a definite integral by the matching u-values, found from the substitution, so the integral can be finished in u.
Integration by parts formula
∫u (dv/dx) dx = uv − ∫v (du/dx) dx (Yes)
Substitution u = g(x)
∫f(g(x)) g′(x) dx = ∫f(u) du; replace dx using du/dx = g′(x). (No — learn it)
∫ln x dx
x ln x − x + c (parts with 1 × ln x). (No — derive it)
∫k/(ax + b) dx
(k/a) ln|ax + b| + c (No — learn it)
∫k/(ax + b)² dx
−k/(a(ax + b)) + c (No — derive it)
∫f′(x)/f(x) dx
ln|f(x)| + c (No — learn it)
What a substitution does to the area under a curve
The region changes shape and width in the u-variable, but the factor from dx = (dx/du) du keeps the area the same.
Area under y = ln x from x = 1 to x = e
Exactly 1: the rectangle e × 1 minus the area e − 1 between the curve and the y-axis.
Vertical asymptotes of y = (5x + 4)/((x − 1)(x + 2))
x = 1 and x = −2, one for each linear factor; a definite integral must not cross them.
Curve near a squared factor (x + 1)² in the denominator
It goes off to infinity in the same direction on both sides of x = −1, unlike a single factor.
How do you choose u in integration by parts?
u should become simpler when differentiated (a power of x), and dv/dx must be integrable. If ln x is present, let u = ln x.
Why does x² eˣ need integration by parts twice?
Each application with u = power of x lowers the power by one; after one use an x eˣ integral remains.
Why must the limits change in a definite substitution?
The limits are values of the variable of integration. Once you integrate with respect to u, they must be u-values.
Why does 1/(x + 1)² not integrate to a logarithm?
Its numerator is not the derivative of its denominator; it is the power (x + 1)⁻², which integrates to −(x + 1)⁻¹.
When is a substitution finished?
When no x and no dx remain: the integrand and the differential are entirely in u.
Which method for ∫x eˣ², ∫x eˣ and ∫3/(x² + 3x)?
Recognition (or u = x²); parts with u = x; partial fractions.
∫x eˣ dx
u = x, v = eˣ: x eˣ − ∫eˣ dx = (x − 1)eˣ + c.
∫x ln x dx
u = ln x, v = x²/2: ½x² ln x − ¼x² + c.
∫x cos 2x dx
u = x, v = ½ sin 2x: ½x sin 2x + ¼ cos 2x + c.
∫x/√(x + 1) dx using u = x + 1
∫(u − 1)u^(−1/2) du = ⅔u^(3/2) − 2u^(1/2) + c = ⅔(x + 1)^(3/2) − 2(x + 1)^(1/2) + c.
∫₀¹ x/(1 + x²)² dx using u = 1 + x²
du = 2x dx, limits 1 to 2: ∫₁² 1/(2u²) du = ½[−1/u]₁² = ¼.
∫3/(x(x + 3)) dx
Partial fractions 1/x − 1/(x + 3): ln|x| − ln|x + 3| + c.
∫₂⁴ (5x + 4)/((x − 1)(x + 2)) dx
3/(x − 1) + 2/(x + 2) → [3 ln(x − 1) + 2 ln(x + 2)]₂⁴ = 3 ln 3 + 2 ln(3/2) = ln(243/4).
∫₁ᵉ ln x dx
[x ln x − x]₁ᵉ = (e − e) − (0 − 1) = 1.
Keeping the x-limits after substituting
Wrong: convert each limit with u = g(x), or substitute back to x before using the old limits.
Writing dx = du after u = 2x − 1
du = 2 dx, so dx = ½ du. Every dx needs its factor.
∫1/(3x + 2) dx = ln|3x + 2| + c?
No: divide by the coefficient of x, ⅓ ln|3x + 2| + c.
Sign in the second application of parts
Bracket it: −[uv − ∫v du] — the minus applies to both terms.
“In the form ln k”
Combine all the log terms into a single logarithm with the laws of logs; keep it exact.
The question says “use the substitution …”
Use exactly that substitution and show the dx link, the new limits and the u-integral.

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