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The product rule integrated, $\int u\,dv = uv - \int v\,du$, and why the choice of $u$ is the whole skill.
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 state integration by parts with the hypotheses the fundamental theorem needs, derive it from the product rule, choose which factor to take as $u$ by asking which one simplifies when differentiated, evaluate the boundary term at both limits rather than assuming it vanishes, keep the sign of the subtraction, and recognise when a trade has gone the wrong way and cost you nothing but one line to undo.
Calculus I ended with substitution, which is the chain rule read backwards: it undoes a composition. The product rule was never undone, and the gap shows up immediately — $\int x e^x\,dx$ is a product of two functions you can each integrate on sight, and no amount of staring produces an answer.
The reason is that integration has no product rule. $\int fg$ is not $\int f \int g$, and there is no formula that turns the integral of a product into anything built from the integrals of its factors. What there is, is a way of trading one integral for another, and that is all integration by parts claims to be.
$u$ is the factor you differentiate; $dv$ is everything else, including the $dx$. $du$ is $u'\,dx$ and $v$ is any antiderivative of $dv$ — any one, since the constant cancels in the same way it does in a definite integral.
The boundary term is $[uv]_a^b$, the part of the answer that is not an integral. The leftover integral is $\int v\,du$, the one you traded for. A reduction formula — the next lesson — is what you get when the leftover integral is the original one with a smaller exponent.
Theorem. If $u$ and $v$ have continuous derivatives on $[a, b]$, then $$\int_a^b u\,v'\,dx = [uv]_a^b - \int_a^b v\,u'\,dx,$$ and in indefinite form $\int u\,dv = uv - \int v\,du$.
Where it comes from. Differentiate a product: $(uv)' = u'v + uv'$. Integrate both sides over $[a,b]$; the left becomes $[uv]_a^b$ by the fundamental theorem, and the right splits by linearity. Rearrange. That is the whole proof, and it is worth knowing because it tells you what the hypotheses are for: the fundamental theorem needs $(uv)'$ continuous.
What it does. It does not evaluate an integral. It trades $\int u\,dv$ for $\int v\,du$, plus a boundary term you get for free. The trade is worth making exactly when $\int v\,du$ is easier than what you started with, and that is decided by your choice of $u$.
Choosing $u$. Take as $u$ the factor whose derivative is simpler than itself: a logarithm, then an inverse trigonometric function, then a polynomial. Leave as $dv$ the factor that is unchanged by either operation — an exponential, a sine, a cosine.
Another way: picture
Draw the curve $v$ against $u$ in the plane and box the rectangle from $(u(a), v(a))$ to $(u(b), v(b))$. The rectangle's area changes by $[uv]_a^b$; the region under the curve is $\int v\,du$ and the region beside it is $\int u\,dv$. The formula is the statement that the two regions make up the change in the rectangle — parts is a picture about splitting a rectangle, and the algebra is bookkeeping on top of it.
Another way: steps
A polynomial times an exponential or a sine. $\int x e^x dx$: take $u = x$. Each pass lowers the degree by one, so a degree-$n$ polynomial needs $n$ passes and then stops. Taking $u = e^x$ instead raises the degree and never stops.
Anything times a logarithm. $\int x \ln x\,dx$: take $u = \ln x$, because $\ln$ has no antiderivative you have met and its derivative is $1/x$, which is algebraic. The same reasoning gives $\int \ln x\,dx$ with $dv = dx$ — the second factor is $1$, and noticing that is the only difficulty in the problem.
A function times itself, twice over. $\int e^x \sin x\,dx$ returns to itself after two passes, with a sign change. You then solve for the integral algebraically rather than continuing. That is the loop case, and it is the subject of the next lesson along with reduction formulas.
Dropping the boundary term. In the tidy examples $[uv]_a^b$ is zero, and the habit of not writing it down survives into the examples where it is not. Write it every time, then evaluate it.
Losing the minus sign. $\int u\,dv = uv - \int v\,du$. The subtraction is part of the formula; a sign error here turns a positive integral negative and nothing in the arithmetic will flag it.
Choosing $u$ to make $v$ pretty. The choice is governed by $du$, not by $v$. A tidy $v$ with a worse $du$ is a trade that loses.
Adding a constant to $v$ and then worrying about it. Any antiderivative works; the extra constant contributes $c[u]_a^b$ to the boundary term and $-c\int u'$ to the leftover, and those cancel exactly.
Expecting it to finish the job. Parts often produces another integral needing substitution, or parts again. It is a step, not an answer.
The formula is often read as though it computed something, so a learner who applies it correctly and is left holding another integral concludes they have done it wrong. They have not. $\int u\,dv = uv - \int v\,du$ is an exchange: you hand over one integral and receive a boundary term plus a different integral. Whether you are better off depends entirely on which factor you called $u$.
That is why the choice is the lesson and the arithmetic is not. A learner who can carry out the algebra but chooses $u$ by which factor is written first will solve the textbook problems, where the first factor happens to be the polynomial, and will fail on the first problem that is written the other way round.
The test is one question: is $\int v\,du$ easier than $\int u\,dv$? If the answer is no, the trade was the wrong way round and it costs one line to swap. A test is its hypotheses. Every convergence test in this course is a theorem with conditions, and a test quoted where its conditions fail has not been applied — it has been guessed with. Naming the hypothesis you checked is part of the answer, not a flourish on top of it.
$\int_0^1 x e^{x}\,dx$: take $u = x$, $dv = e^x dx$, so $du = dx$ and $v = e^x$.
The polynomial is the factor that simplifies.
$= [x e^x]_0^1 - \int_0^1 e^x dx = e - [e^x]_0^1 = e - (e - 1) = 1$.
Boundary term first, then the leftover.
Had we taken $u = e^x$ and $dv = x\,dx$, the leftover would have been $\int \tfrac{x^2}{2}e^x dx$ — the same shape with a higher power. That is what a losing trade looks like.
$\int_1^{e} \ln x\,dx$ has only one factor, so the other one is $1$: $u = \ln x$, $dv = dx$.
The trick is seeing that $1$ is a factor.
$du = \dfrac{dx}{x}$, $v = x$, so the integral is $[x\ln x]_1^e - \int_1^e x \cdot \dfrac{dx}{x} = e - \int_1^e dx = e - (e - 1) = 1$.
The leftover collapses to something elementary.
The hypotheses hold: $\ln x$ and $x$ have continuous derivatives on $[1, e]$. On $[0, e]$ they would not, and the integral would be improper — a different question, handled in unit 3.
Take $u = x$, $dv = (x + 1)^4 dx$, so $du = dx$ and $v = \dfrac{(x+1)^5}{5}$.
The polynomial factor of degree one is the one that simplifies.
Boundary term: $\left[\dfrac{x(x+1)^5}{5}\right]_0^1 = \dfrac{32}{5}$.
Both ends, honestly — the lower one is zero, the upper one is not.
Leftover: $\displaystyle\int_0^1 \dfrac{(x+1)^5}{5}dx = \dfrac{63}{30} = \dfrac{21}{10}$. The answer is $\dfrac{32}{5} - \dfrac{21}{10} = \dfrac{43}{10}$.
Evaluate $\displaystyle\int_0^1 x(1 - x)^{6}\,dx$.
The integral equals answer.
For $\int x^2 \sin x\,dx$, which kind of factor should be taken as $u$?
Put the steps of evaluating $\displaystyle\int_0^1 x(1 - x)^{4}dx$ by parts into the order you do them.
Number the steps in order (write the number in the box):
Fill in the two terms of $\displaystyle\int_0^1 x(1 - x)^{7}dx = [uv]_0^1 - \int_0^1 v\,du$, with $u = x$ and $v = -\dfrac{(1 - x)^{8}}{8}$.
| Value | |
|---|---|
| The boundary term $[uv]_0^1$ | 0 |
| The leftover integral $\int_0^1 v\,du$ | |
| The value of the original integral |
Evaluate $\displaystyle\int_{3}^{4} x\,(x - 3)^{4}\,dx$.
The integral equals answer.
$f$ has a continuous derivative on $[0, 1]$, with $f(1) = 11$ and $\displaystyle\int_0^1 f(x)\,dx = 5$. Find $\displaystyle\int_0^1 x\,f'(x)\,dx$.
The integral equals answer.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Build the derivation of $\displaystyle\int_a^b u\,v'\,dx = [uv]_a^b - \int_a^b v\,u'\,dx$ from the product rule.
This task has no paper form; do it on a device.
You can carry out a by-parts calculation and, more importantly, say before you start whether the trade is worth making. Say in your own words why the formula is an exchange rather than an evaluation. Next: what to do when the exchange hands you back the integral you started with.
10. Your turn: $\displaystyle\int_0^{1} x\,(x + 1)^{4}\,dx$, step 3
Subtract, and keep the sign.