Corrected exercises: substitution, integration by parts and partial fractions (201-SN3-RE)

Form $\int\frac{f'}{f}$, whose antiderivative is $\ln\left|f\right|$. The numerator $3x^{2}+2$ is exactly the derivative of $x^{3}+2x+1$: the answer is $\ln\left|x^{3}+2x+1\right|+C$. Likewise $\cos x$ is the derivative of $2+\sin x$, hence $\ln\left(2+\sin x\right)+C$, the absolute value being unnecessary since $2+\sin x$ stays positive.

Form $\int f'f^{n}$, with antiderivative $\frac{f^{n+1}}{n+1}$. Here $x$ is half the derivative of $x^{2}+1$: $\frac{1}{2}\int u^{5}du=\frac{\left(x^{2}+1\right)^{6}}{12}+C$. For the second, $\frac{1}{x}$ is the derivative of $\ln x$, hence $\frac{\left(\ln x\right)^{4}}{4}+C$.

Form $\int f'e^{f}$, with antiderivative $e^{f}$. Since $\cos x=\left(\sin x\right)'$: $e^{\sin x}+C$. Since $\sec^{2}x=\left(\tan x\right)'$: $e^{\tan x}+C$. These two integrals look fearsome and are settled without writing a single line of substitution, provided you have the table of derivatives in your head.

Form $\int\frac{f'}{\sqrt{f}}$, with antiderivative $2\sqrt{f}$. Here $x=\frac{1}{2}\left(x^{2}+4\right)'$, hence $\sqrt{x^{2}+4}+C$. For the second, $\frac{1}{2\sqrt{x}}$ is the derivative of $1+\sqrt{x}$: the integral equals $2\int\frac{du}{u^{2}}$ with $u=1+\sqrt{x}$, that is $-\frac{2}{1+\sqrt{x}}+C$.

The five forms: $\int\frac{f'}{f}=\ln\left|f\right|$; $\int f'f^{n}=\frac{f^{n+1}}{n+1}$ for $n\neq -1$; $\int f'e^{f}=e^{f}$; $\int\frac{f'}{\sqrt{f}}=2\sqrt{f}$; and $\int f'\cos f=\sin f$, with its trigonometric variants. The test fits in one question: mentally DIFFERENTIATE the most complicated expression in the integrand, the one inside a power, a logarithm, an exponential or a root, and look at whether the result appears elsewhere in the integrand, up to a constant factor. If it does, the substitution is immediate; if not, no direct substitution will work and you must change technique.

Let $u=x+1$, so $x=u-1$ and $dx=du$: $\int(u-1)\sqrt{u}\,du=\int\left(u^{3/2}-u^{1/2}\right)du=\frac{2u^{5/2}}{5}-\frac{2u^{3/2}}{3}$, that is $\frac{2\left(x+1\right)^{5/2}}{5}-\frac{2\left(x+1\right)^{3/2}}{3}+C$. The remaining $x$ was replaced by $u-1$: that is the whole trick.

$u=x+2$, $x=u-2$: $\int\frac{u-2}{u^{3}}du=\int\left(u^{-2}-2u^{-3}\right)du=-\frac{1}{u}+\frac{1}{u^{2}}$, that is $-\frac{1}{x+2}+\frac{1}{\left(x+2\right)^{2}}+C$.

$u=x^{2}+1$ gives $du=2x\,dx$: one factor $x$ is ABSORBED by the $du$, and what remains is $x^{2}=u-1$, which is expressed without a root. $\frac{1}{2}\int(u-1)\sqrt{u}\,du=\frac{u^{5/2}}{5}-\frac{u^{3/2}}{3}$, that is $\frac{\left(x^{2}+1\right)^{5/2}}{5}-\frac{\left(x^{2}+1\right)^{3/2}}{3}+C$. The difference from the first two cases is that the power of $x$ was ODD: one factor goes into the $du$ and the rest, an even power, converts cleanly. This is exactly the mechanism behind odd powers of sine.

$u=\sqrt{x}$, so $x=u^{2}$ and $dx=2u\,du$: $\int\frac{2u}{1+u}du=2\int\left(1-\frac{1}{1+u}\right)du=2u-2\ln\left|1+u\right|$, that is $2\sqrt{x}-2\ln\left(1+\sqrt{x}\right)+C$. Note the division done along the way: as soon as a quotient has comparable degrees, divide before integrating.

In the second one, the derivative of the inside is $2x$: the factor $x$ in front of the root IS the hidden derivative, the substitution absorbs everything and you get $\frac{\left(x^{2}+1\right)^{3/2}}{3}+C$ in two lines. In the first one, the derivative of $x+1$ is 1: the factor $x$ has nothing to do with it, it survives the substitution and must be re-expressed, which doubles the length of the calculation. The general rule is that the cost of a substitution is measured by what it leaves standing.

$u=e^{x}$ gives $du=e^{x}dx$, so $dx=\frac{du}{u}$: $\int\frac{du}{u\left(1+u\right)}$. The decomposition $\frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}$ gives $\ln\left|u\right|-\ln\left|1+u\right|=\ln\frac{u}{1+u}$, that is $x-\ln\left(1+e^{x}\right)+C$. An exponential integral has become a rational function in two lines.

$u=e^{x}$ and $dx=\frac{du}{u}$: $\int\frac{u^{2}}{1+u}\cdot\frac{du}{u}=\int\frac{u}{1+u}du=\int\left(1-\frac{1}{1+u}\right)du=u-\ln\left|1+u\right|$, that is $e^{x}-\ln\left(1+e^{x}\right)+C$.

You need a substitution that makes BOTH exponents whole numbers at once: let $x=u^{6}$, because 6 is the least common multiple of 2 and 3. Then $\sqrt{x}=u^{3}$, $\sqrt[3]{x}=u^{2}$ and $dx=6u^{5}du$: $\int\frac{6u^{5}}{u^{3}+u^{2}}du=\int\frac{6u^{3}}{u+1}du$. Long division gives $\frac{u^{3}}{u+1}=u^{2}-u+1-\frac{1}{u+1}$, hence $2u^{3}-3u^{2}+6u-6\ln\left|u+1\right|+C$ with $u=x^{1/6}$. A smaller exponent would leave a root, and a larger one would make the polynomial needlessly heavy.

Multiply top and bottom by $e^{x}$: $\int\frac{e^{x}dx}{e^{2x}+1}$. With $u=e^{x}$, this is $\int\frac{du}{u^{2}+1}=\arctan u$, that is $\arctan\left(e^{x}\right)+C$. The preliminary multiplication was essential: without it, no substitution gives a clean result. Notice along the way that $\frac{1}{e^{x}+e^{-x}}=\frac{1}{2\cosh x}$, which explains why this result is related to the hyperbolic functions.

In front of $\sqrt[n]{x}$, let $x=u^{n}$. In front of several roots with different indices, let $x=u^{m}$ where $m$ is the LCM of the indices. In front of $e^{x}$, let $u=e^{x}$, which always gives $dx=\frac{du}{u}$ and turns any fraction in $e^{x}$ into a rational function of $u$. In front of $\sqrt[n]{ax+b}$, let $u=\sqrt[n]{ax+b}$, so $x=\frac{u^{n}-b}{a}$. In all four cases the goal is the same: make the fractional exponents disappear to land on a rational function, the only family we know how to integrate systematically.

$u=\ln x$, $du=\frac{dx}{x}$; $x=1$ gives $u=0$ and $x=e$ gives $u=1$: $\int_{0}^{1}u\,du=\frac{1}{2}$.

For the first, $u=x^{2}+1$ goes from 1 to 2: $\frac{1}{2}\int_{1}^{2}\frac{du}{u}=\frac{\ln 2}{2}\approx 0.3466$. For the second, $u=e^{x}$ goes from 1 to 2: $\int_{1}^{2}\frac{du}{1+u^{2}}=\arctan 2-\arctan 1\approx 1.1071-0.7854\approx 0.3218$.

$u=\sqrt{x}$ goes from 0 to 2, with $dx=2u\,du$: $\int_{0}^{2}\frac{2u}{1+u}du=2\int_{0}^{2}\left(1-\frac{1}{1+u}\right)du=2\left[u-\ln\left(1+u\right)\right]_{0}^{2}=2\left(2-\ln 3\right)\approx 1.803$.

The answer is RIGHT, but the reasoning is wrong, and that is worse. The integrand $x^{3}\left(x^{4}+1\right)^{2}$ is an ODD function integrated over a symmetric interval: that is why it equals zero. The argument of equal limits is worthless here: the substitution $u=x^{4}+1$ is not one-to-one on $\left[-1, 1\right]$, since $x$ and $-x$ give the same $u$. Writing $\int_{2}^{2}$ amounts to folding the interval onto itself, merging two pieces that do cancel here, but that would ADD UP for an even function. The same reasoning applied to $\int_{-1}^{1}x^{3}\left(x^{4}+1\right)^{2}\cdot x\,dx$, an even integrand that is positive everywhere except at 0, would again give zero, which would be absurd.

The substitution must be MONOTONIC, hence one-to-one, on the interval of integration: that is what guarantees a correspondence without folding between the values of $x$ and those of $u$. Counterexample: on $\left[-1, 1\right]$, the substitution $u=x^{2}$ fails, because it sends $-1$ and $1$ to the same $u=1$. For $\int_{-1}^{1}x^{2}dx$, it would give $\int_{1}^{1}$, so zero, whereas the true value is $\frac{2}{3}$. The remedy is to split the interval at $x=0$, where the substitution stops being monotonic, and to handle the two pieces separately.

The first contains the factor $x$, half the derivative of $x^{2}$: immediate substitution, $\frac{e^{x^{2}}}{2}+C$. The second does not, and no substitution can handle it. Better still: its antiderivative is not an ELEMENTARY function, it cannot be expressed with any finite combination of the usual functions. This is not a limit of our skill but a theorem, and the only acceptable answer is numerical, or by a power series.

The first has the form $\int f'f$ with $f=\ln x$: it equals $\frac{\left(\ln x\right)^{2}}{2}+C$. The second has no factor $\frac{1}{x}$: it calls for integration by PARTS, with the invisible factor 1, and gives $x\ln x-x+C$. The presence or absence of the $\frac{1}{x}$ tips you from one technique to the other.

With the factor $x$: $-\sqrt{1-x^{2}}+C$ by substitution, and $\frac{1}{2}\ln\left|x^{2}-1\right|+C$ likewise. Without it: $\arcsin x+C$, which requires a TRIGONOMETRIC substitution, and $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right|+C$, which requires PARTIAL FRACTIONS. A single factor $x$ therefore separates four different techniques, and it is the first detail to look at in the whole chapter.

That the degree of the numerator EQUALS that of the denominator: the fraction is improper, and no substitution can apply before the division. $\frac{x^{2}+1}{x^{2}-4}=1+\frac{5}{x^{2}-4}$, and the remainder is decomposed into partial fractions. The rule is absolute: as soon as a rational function has a numerator of degree greater than or equal to that of the denominator, LONG DIVISION comes before everything else.

First signal: the integrand is a PRODUCT of two functions from different families, a polynomial and an exponential, a polynomial and a logarithm, an inverse trigonometric function and something else, without one being the derivative of the inside of the other. That is integration by parts. Second signal: the integrand is a RATIONAL FUNCTION, a quotient of two polynomials, without the numerator being proportional to the derivative of the denominator. That is partial fractions, preceded by a division if the degree of the top reaches that of the bottom. Third signal: a ROOT of the form $\sqrt{a^{2}\pm x^{2}}$ or $\sqrt{x^{2}-a^{2}}$ remains, with no factor $x$ to absorb it. That is trigonometric substitution. And if none of the three applies and substitution fails too, consider that the antiderivative may simply not exist in elementary form.

The logarithm is at the front of LIATE: take $u=\ln x$ and $dv=x\,dx$, so $du=\frac{dx}{x}$ and $v=\frac{x^{2}}{2}$. Hence $\frac{x^{2}}{2}\ln x-\int\frac{x^{2}}{2}\cdot\frac{dx}{x}=\frac{x^{2}}{2}\ln x-\frac{1}{2}\int x\,dx=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C$. The remaining integral has become a polynomial one: the logarithm has disappeared, which is exactly what we wanted.

Here A comes before E: $u=x$ and $dv=e^{-x}dx$, so $v=-e^{-x}$. $\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx=-xe^{-x}-e^{-x}=-\left(x+1\right)e^{-x}+C$. Likewise, $u=x$ and $dv=\cos x\,dx$ give $x\sin x-\int\sin x\,dx=x\sin x+\cos x+C$. In both cases, differentiating $x$ makes it disappear.

There is only one factor: to apply the formula you MUST make a second one, and the only one available is the factor 1, invisible in the notation. So let $u=\ln x$ and $dv=1\cdot dx$, hence $v=x$: $x\ln x-\int x\cdot\frac{dx}{x}=x\ln x-x+C$. This is not a trick but the only possible route, because no antiderivative of the logarithm is known when you start, which rules out taking it as $dv$.

With $u=e^{x}$ and $dv=x\,dx$: $v=\frac{x^{2}}{2}$ and $du=e^{x}dx$, so $\int xe^{x}dx=\frac{x^{2}}{2}e^{x}-\frac{1}{2}\int x^{2}e^{x}dx$. The equality is perfectly TRUE, but useless: the degree of the polynomial went from 1 to 2, the remaining integral is harder than the original, and repeating the operation would only make it worse indefinitely. Integration by parts is never wrong; it is only fruitless or productive.

The functions at the front, logarithms and inverse trigonometric functions, have two properties: their DERIVATIVE is algebraic, hence simpler than they are, and their antiderivative is not known at the start. They are therefore the obvious candidates for $u$, the role of the one you differentiate. The functions at the back, exponentials and trigonometric functions, have the opposite property: they reproduce themselves when integrated, without getting more complicated, which makes them convenient choices for $dv$. Algebraic functions sit in the middle because they lose one degree at each differentiation: they eventually disappear, provided you differentiate them. LIATE is therefore not an arbitrary recipe, it is the ranking of functions according to what differentiation does to them.

First round: $u=x^{2}$, $dv=e^{x}dx$, hence $x^{2}e^{x}-2\int xe^{x}dx$. Second round on the remaining integral: $\int xe^{x}dx=xe^{x}-e^{x}$. Altogether, $x^{2}e^{x}-2xe^{x}+2e^{x}+C=e^{x}\left(x^{2}-2x+2\right)+C$. The polynomial lost one degree at each round, until it disappeared.

Draw two columns. On the left, the ALGEBRAIC factor, which you DIFFERENTIATE from row to row until you get zero. On the right, the other factor, which you INTEGRATE from row to row as many times. Then multiply along the diagonals, each product taking an alternating sign $+$, $-$, $+$, $-$, starting with $+$. Stop as soon as the left column reaches zero, which happens after $n+1$ rows for a polynomial of degree $n$.

Left column: $x^{3}$, $3x^{2}$, $6x$, $6$, $0$. Right column: $e^{2x}$, $\frac{e^{2x}}{2}$, $\frac{e^{2x}}{4}$, $\frac{e^{2x}}{8}$, $\frac{e^{2x}}{16}$. Diagonal products, with alternating signs: $+x^{3}\frac{e^{2x}}{2}-3x^{2}\frac{e^{2x}}{4}+6x\frac{e^{2x}}{8}-6\frac{e^{2x}}{16}$, that is $e^{2x}\left(\frac{x^{3}}{2}-\frac{3x^{2}}{4}+\frac{3x}{4}-\frac{3}{8}\right)+C$. Three integrations by parts carried out in one go.

Left column: $x^{2}$, $2x$, $2$, $0$. Right column: $\sin 3x$, $-\frac{\cos 3x}{3}$, $-\frac{\sin 3x}{9}$, $\frac{\cos 3x}{27}$. Hence $-\frac{x^{2}\cos 3x}{3}+\frac{2x\sin 3x}{9}+\frac{2\cos 3x}{27}+C$, remembering that the alternating sign of the third term, $+$, combines with the sign of the antiderivative.

For $\int x\ln x\,dx$, NO: the column to differentiate would have to contain the logarithm, which never reaches zero by differentiation, and the column to integrate would contain $x$, whose degree would grow indefinitely. The tabular method requires a polynomial factor that eventually becomes zero. For $\int e^{x}\cos x\,dx$, NO as well, but for another reason: neither factor becomes zero by differentiation, so the two columns would go on forever. This is the CYCLIC case, which is solved differently, as the next exercise shows.

Let $I=\int e^{x}\cos x\,dx$. With $u=e^{x}$ and $dv=\cos x\,dx$: $I=e^{x}\sin x-\int e^{x}\sin x\,dx$. On the latter, the same choice: $\int e^{x}\sin x\,dx=-e^{x}\cos x+\int e^{x}\cos x\,dx=-e^{x}\cos x+I$. Substituting back: $I=e^{x}\sin x+e^{x}\cos x-I$, so $2I=e^{x}\left(\sin x+\cos x\right)$ and $I=\frac{e^{x}\left(\sin x+\cos x\right)}{2}+C$.