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Any quadratic becomes $au^2 + q$ after a shift, and a root of a linear expression becomes rational when you substitute for the root itself.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to complete the square in any quadratic, including one whose leading coefficient is not $1$, identify from the sign of the leftover constant which of the three standard forms you have and therefore which integral or substitution applies, take the square root of the leftover rather than the leftover itself as $a$, split a linear numerator against the derivative of the denominator before shifting, move the limits of a definite integral with the shift, and rationalise a root of a linear expression by substituting for the root.
The last lesson gave three radicals — $\sqrt{a^2 - x^2}$, $\sqrt{a^2 + x^2}$, $\sqrt{x^2 - a^2}$ — and a substitution for each. Almost no quadratic you actually meet is written that way. $\sqrt{x^2 + 6x + 13}$ has a linear term, and none of the three substitutions touches it.
The fix is not a fourth substitution. It is the observation that every quadratic is one of those three after a shift, and finding the shift is ordinary school algebra: complete the square.
To complete the square in $x^2 + bx + c$ is to write it as $\left(x + \tfrac{b}{2}\right)^2 + \left(c - \tfrac{b^2}{4}\right)$. The number inside the bracket is the shift; the second bracket is the leftover constant, and its sign decides which standard form you have.
The substitution $u = x + \tfrac{b}{2}$ is a translation: $du = dx$, so nothing has to be divided by anything, and the only thing that changes in a definite integral is the limits.
A rationalising substitution is a different preparation for a different obstacle: when a root of a linear expression appears, setting $u$ equal to the root itself removes it, because $u^2$ is then a polynomial.
Completing the square. $$ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right).$$ Factor out the leading coefficient first, halve what is left multiplying $x$, square it, and correct the constant.
With $u = x + \dfrac{b}{2a}$ the expression becomes $au^2 + q$, which is one of
A linear numerator. $\displaystyle\int\frac{mx + n}{x^2 + bx + c}dx$ is split rather than shifted: write the numerator as a multiple of the derivative of the denominator plus a constant. The first piece integrates to a logarithm by substitution; the second is one of the three forms above. Doing it in that order is what keeps it to two lines.
Rationalising substitutions. If the integrand contains $\sqrt[n]{ax + b}$, set $u = \sqrt[n]{ax + b}$. Then $u^n = ax + b$, $x$ is a polynomial in $u$, and $dx$ is too, so the whole integrand becomes rational. For $\sqrt{x}$ and $\sqrt[3]{x}$ together, take $u = x^{1/6}$ — the least common multiple of the roots.
Another way: picture
The graph of $y = x^2 + bx + c$ is the parabola $y = x^2$ moved: left by $b/2$, and up or down by the leftover constant. Completing the square is nothing but reading those two movements off the coefficients. The substitution $u = x + b/2$ slides the picture back so the vertex sits on the axis, which is where all the standard formulas are written.
Another way: steps
$\displaystyle\int\frac{3x + 1}{x^2 + 4x + 13}\,dx$ looks like it needs both techniques at once, and it does — but in a fixed order.
The derivative of the denominator is $2x + 4$. Write the numerator as a multiple of it plus a constant: $3x + 1 = \tfrac32(2x + 4) - 5$. Then the integral splits as $$\frac32\int\frac{2x+4}{x^2+4x+13}dx - 5\int\frac{dx}{x^2+4x+13}.$$
The first is $\tfrac32\ln(x^2 + 4x + 13)$ by the substitution $w = x^2 + 4x + 13$ — the numerator is exactly $dw$. The second is where completing the square comes in: $x^2 + 4x + 13 = (x+2)^2 + 9$, so it is $-\tfrac53\arctan\dfrac{x+2}{3}$.
The order matters because completing the square first leaves a numerator of $3u - 5$, which then has to be split anyway — the same work, done after an unnecessary change of variable. Split first, shift second.
Forgetting to factor out the leading coefficient. In $3x^2 + 12x + 5$ the shift is $2$, not $6$: halve the coefficient of $x$ after the $3$ has come out.
Correcting the constant with the wrong sign. The bracket adds the square, so the constant must lose it. Writing $(x+3)^2 + 9$ for $x^2 + 6x + 9$ gives an expression that is $9$ too big everywhere.
Taking the leftover as $a$ rather than $a^2$. $(x+2)^2 + 9$ has $a = 3$. The arctangent formula carries $1/a$ and $u/a$, so the error appears twice.
Not moving the limits. A definite integral substituted with $u = x + p$ runs from $\alpha + p$ to $\beta + p$. The shift is invisible in $du$ and easy to forget for that reason.
Ignoring that $u^2 - a^2$ factors. Under a root it is a secant substitution; in a denominator it is two linear factors and partial fractions, which is far shorter. Choosing by habit rather than by where the quadratic sits costs a page.
Rationalising a root of something that is not linear. $u = \sqrt{x^2 + 1}$ does not make the integrand rational, because solving for $x$ reintroduces a root. The technique is for roots of linear expressions.
Completing the square is taught in school and feels like something beneath a calculus course, so it gets done hurriedly or skipped in favour of hunting for the right substitution. That inverts the difficulty. By the time the quadratic is in standard form the rest of the problem is a table lookup; before it is, no table applies and no amount of cleverness with substitutions helps.
The sign of the leftover constant is not a detail either — it decides which of three entirely different answers you are heading for. $(x+2)^2 + 9$ gives an arctangent. $(x+2)^2 - 9$ factors and gives logarithms. $9 - (x+2)^2$ gives an arcsine. Three quadratics that differ only in signs produce three unrelated functions, and the only place that difference is visible is in the completed square.
So the useful habit is to finish the algebra before choosing the calculus. Write the quadratic in standard form, look at the two signs, and only then decide what kind of integral you are doing.
$\displaystyle\int\frac{dx}{x^2 - 6x + 25}$: the denominator is $(x - 3)^2 + 16$.
Halve $-6$ to get the shift $-3$; $25 - 9 = 16$ is left.
Substitute $u = x - 3$, $du = dx$: $\displaystyle\int\frac{du}{u^2 + 16}$, which is $\dfrac14\arctan\dfrac u4 + C$.
The leftover is $16$, so $a = 4$.
Back in $x$: $\dfrac14\arctan\dfrac{x - 3}{4} + C$. Differentiating returns the integrand, which takes ten seconds and is the only check available.
$\displaystyle\int_0^{3}\frac{dx}{1 + \sqrt{x + 1}}$: put $u = \sqrt{x+1}$, so $u^2 = x + 1$ and $2u\,du = dx$.
Set $u$ equal to the root itself.
The limits go with it: $x = 0$ gives $u = 1$, $x = 3$ gives $u = 2$. The integral is $\displaystyle\int_1^2\frac{2u\,du}{1 + u}$ — rational, with no root anywhere.
Both $x$ and $dx$ become polynomials in $u$.
Divide: $\dfrac{2u}{1+u} = 2 - \dfrac{2}{1+u}$, so the integral is $\left[2u - 2\ln(1+u)\right]_1^2 = 2 - 2\ln\dfrac32$.
An improper fraction is divided before anything else.
Factor the $-1$ out of the $x$ terms: $8 + 2x - x^2 = 8 - (x^2 - 2x)$.
The leading coefficient is $-1$, so take it out first.
$x^2 - 2x = (x-1)^2 - 1$, so the whole thing is $9 - (x - 1)^2$.
Halve $-2$; correct the constant.
With $u = x - 1$ this is $\displaystyle\int\frac{du}{\sqrt{9 - u^2}} = \arcsin\dfrac{u}{3} + C = \arcsin\dfrac{x-1}{3} + C$.
Write $x^{2} + 14x + 50$ in the form $(x + p)^{2} + q$.
Answer:
With the substitution $u = x + 4$, the integral $\displaystyle\int\dfrac{dx}{x^{2} + 8x + 80}$ becomes $\displaystyle\int\dfrac{du}{u^{2} + a^{2}}$. Find $a$.
The standard-form parameter is a = answer.
Completing the square turns $x^{2} + 2x - 24$ into $(x + 1)^{2} - 25$. Which standard form is that, in $u = x + 1$?
Complete the square in $x^{2} + 12x + 45$ by filling in the three numbers it takes.
| Value | |
|---|---|
| The coefficient of $x$ | 12 |
| Half of it, which goes inside the bracket | |
| Its square, which the bracket adds | |
| The constant left over |
For which $x$ is $\sqrt{25 - (x - 4)^{2}}$ real? Give the interval.
This task has no paper form; do it on a device.
A stone thrown upward has height $h(t) = 42t - 3t^{2}$ metres at time $t$. What is its greatest height?
The greatest height is answer metres.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Write $x^{2} + 6x + 11$ in the form $(x + p)^{2} + q$.
Answer:
You can put any quadratic into standard form and say which of three different answers it is heading for. Say in your own words why the sign of the leftover constant matters more than its size. Next: what to do with a rational function whose denominator factors.
10. Your turn: $\displaystyle\int\frac{dx}{\sqrt{8 + 2x - x^{2}}}$, step 3
Constant minus square: the sine form.