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The product rule read backwards: $\int u\,dv = uv - \int v\,du$, the choice of $u$ that makes the trade worth making, and the definite-integral form where the $uv$ term often vanishes.
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 can integrate a product by parts, choosing $u$ to be the factor that gets simpler when differentiated and $dv$ to be the one you can integrate. You can carry the limits of a definite integral through the $uv$ term as well as the new integral, recognise a logarithm as a factor that must be $u$, and tell a product that wants parts from one that is really a substitution in disguise.
You can integrate by substitution, which is the chain rule read backwards. Parts is the product rule read backwards, and it is the technique for the integrals substitution cannot touch: a product of two factors that have nothing to do with each other, like $x$ and $e^{x}$.
Integration by parts: $\int u\,dv = uv - \int v\,du$.
$u$: the factor you differentiate — chosen because differentiating it makes it simpler.
$dv$: the rest of the integrand, including the $dx$ — chosen because you can integrate it.
The new integral: $\int v\,du$, which is what you have traded for, and which has to be easier or the trade was bad.
Differentiating a product gives $(uv)' = u'v + uv'$. Integrate both sides and rearrange:
$$\int u\,dv = uv - \int v\,du$$
That is the whole theorem. It does not evaluate an integral; it exchanges one integral for another, and the technique is worth using exactly when the new one is easier than the old one.
So everything turns on the choice of $u$. Take $u$ to be the factor that gets simpler when you differentiate it, and $dv$ to be the one you can integrate. In $\int x e^{x}dx$, differentiating $x$ gives $1$ and the new integral is $\int e^{x}dx$, which is finished. Choosing $u = e^{x}$ instead gives $\int \tfrac{x^2}{2}e^{x}dx$ — true, legal, and worse.
A logarithm is always $u$. There is nothing else to do with $\ln x$: it has no elementary antiderivative you know yet, and differentiating it gives $\tfrac1x$. That is how $\int \ln x\,dx$ is done at all, with $dv = dx$ and $v = x$.
A definite integral carries the limits through both terms, including the $uv$ term. Evaluate that term at both ends before doing anything else: it often vanishes, and when it does the rest of the problem is one integral.
Another way: picture
Think of the integrand as a load carried by two people. Parts lets you hand the whole of the differentiation to one of them and the whole of the integration to the other. You choose who gets which, and you choose so that the person who has to differentiate is holding the thing that falls apart when differentiated.
Another way: steps
| The integrand contains | Take $u$ to be | Because |
|---|---|---|
| A logarithm | the logarithm | $\ln x$ differentiates to $\tfrac1x$ |
| A power of $x$ times $e^{x}$ | the power | each round drops the power by one |
| A power of $x$ times $\sin x$ | the power | same reason |
| $\ln x$ alone | $\ln x$, with $dv = dx$ | there is nothing else to choose |
The one thing never to take as $u$ is an exponential or a trigonometric function when the other factor is a power: those do not simplify, so the trade gains nothing.
Forgetting the minus sign. The formula is $uv$ minus the new integral, and the minus belongs to the whole of it, not to its first term.
Leaving the $dx$ out of $dv$. $dv$ is $\sin x\,dx$, not $\sin x$. Written without it, the next step invites you to differentiate when you meant to integrate.
Reaching for parts on a product that is a substitution. $\int x e^{x^2}dx$ looks like $\int x e^{x}dx$ and is nothing like it: the $x$ outside is the derivative of the inside, so it is a substitution. Ask that question first, every time.
$\int x e^{x}dx$: take $u = x$ and $dv = e^{x}dx$, so $du = dx$ and $v = e^{x}$.
The power simplifies.
$\int u\,dv = uv - \int v\,du = x e^{x} - \int e^{x}dx$.
One trade.
$= x e^{x} - e^{x} + C$. The new integral was finished on sight.
$\int \ln x\,dx$ is not a product until you make it one: take $u = \ln x$ and $dv = dx$.
There is nothing else to choose.
$du = \tfrac1x dx$ and $v = x$, so the formula gives $x\ln x - \int x\cdot\tfrac1x dx$.
The $x$ cancels.
$= x\ln x - x + C$. Check it by differentiating: $\ln x + 1 - 1 = \ln x$.
Take $u = x$ and $dv = \cos x\,dx$, since the power is what simplifies.
Choose first.
Then $du = dx$ and $v = \sin x$, so the formula gives $x\sin x - \int \sin x\,dx$.
Assemble.
$= x\sin x + \cos x + C$ — and note that the minus sign in the formula and the minus from integrating the sine have cancelled.
Match each integral to the factor that should be $u$.
| $u = x$ | $u = \ln x$, with $dv = x\,dx$ | $u = x^2$ | $u = \ln x$, with $dv = dx$ | |
|---|---|---|---|---|
| $\int x e^{x}\,dx$ | ||||
| $\int x\ln x\,dx$ | ||||
| $\int x^2\cos x\,dx$ | ||||
| $\int \ln x\,dx$ |
To evaluate $\int x^2 e^{x}\,dx$ by parts, what should $u$ be?
Put the steps of evaluating $\int 4x e^{x}\,dx$ by parts into the order you do them.
Number the steps in order (write the number in the box):
By parts, $\int x\sin(5x)\,dx = -\dfrac{x\cos(5x)}{A} + \dfrac{\sin(5x)}{B} + C$. Give $A$, then $B$.
A = p, B = q
Evaluate $\displaystyle\int_0^1 x(1 - x)^{5}\,dx$.
Answer:
Select every integral that integration by parts is the right first tool for.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Put the steps of evaluating $\int 4x e^{x}\,dx$ by parts into the order you do them.
Number the steps in order (write the number in the box):
You can integrate by parts and choose $u$ deliberately. Without looking: what is the formula, which factor should be $u$ in a product with a logarithm in it, and what does a new integral that is harder than the old one tell you?
9. Your turn: $\int x\cos x\,dx$, step 3