Choose the parts of the integral
Two regions beside one curve
The integrand of \(\int x e^x\,dx\) is a product of two factors, \(x\) and \(e^x\). Integration by parts changes this integral into a new one, and you choose how.
The factor you differentiate is called \(u\). The rest of the integrand, with \(dx\), is called \(dv\), and you integrate it to find \(v\). Here \(u = x\) and \(dv = e^x\,dx\), so \(v = e^x\).
The axis across is \(u\) and the axis up is \(v\). Drag the orange point along the curve \((u, v) = (x, e^x)\) as \(x\) runs from 0 to 1.
The green region between the curve and the \(v\)-axis has the area \(\int u\,dv = \int x e^x\,dx\), the integral you started with. The teal region under the curve has the area \(\int v\,du = \int e^x\,dx\), the new integral.
Press Swap the parts to make \(u = e^x\) and \(dv = x\,dx\). The new integral becomes \(\int \tfrac12x^2e^x\,dx\), and its region turns red: the power of \(x\) went up from 1 to 2, so this integral is harder.
Any product
Any product \(f(x)\,g(x)\) can be split in two ways, because either factor can be \(u\), the one you differentiate. The green region is still the integral you started with, and the teal region is the new one.
Press Choose the other split. The product stays the same, but the curve and the new integral change.
A split is good when the new integral is easier than the original one, so the teal region turns red when it is harder.
Notation
Split the integrand into \(u\), the part you differentiate, and \(dv\), the part with \(dx\) that you integrate.
\[\cg{\int u\,dv}\]Find \(du\) by differentiating \(u\), and find \(v\) by integrating \(dv\). Any antiderivative works, so you do not add \(+C\) here. The thin teal strip is \(v\) tall and \(du\) wide, so the teal region is \(\int v\,du\). The thin green strip is \(u\) long and \(dv\) tall, so the green region is \(\int u\,dv\).
\[du = u'\,dx, \qquad v = \int dv\]The two regions fill the orange rectangle, whose area is \(uv\). In words, the integral of \(u\,dv\) equals the product \(uv\) minus the new integral of \(v\,du\).
\[\cg{\int u\,dv} = \cy{uv} - \cb{\int v\,du}\]Choose \(dv\) as a part that you can integrate. Choose \(u\) as a part that becomes simpler when you differentiate it.
Differentiating lowers a power of \(x\) by one and turns \(\ln x\) into \(1/x\). The functions \(e^{kx}\), \(\sin kx\) and \(\cos kx\) are simple to integrate, so they usually go in \(dv\). A factor with no basic antiderivative must be \(u\).
Test the choice before you continue: it works if \(\int v\,du\) is easier than the original integral. The teal region turns red for a moment to show a choice that fails.
When it fails
Consider \(\int x^3 e^{x^2}\,dx\). If you choose \(dv = e^{x^2}\,dx\), you cannot find \(v\), because \(e^{x^2}\) has no basic antiderivative. The curve of \(v\) against \(u\) cannot be drawn.
Keep one factor of \(x\) with the exponential. Then \(dv = x e^{x^2}\,dx\), and the substitution \(w = x^2\) gives \(v = \tfrac12 e^{x^2}\). With \(u = x^2\), the new integral is \(\int x e^{x^2}\,dx\), and the same substitution finds it.
Example 1: A logarithm must be u
Choose \(u\) and \(dv\) for \(\displaystyle\int \sqrt{x}\,\ln x\,dx\). Then find the answer.
The function \(\ln x\) has no basic antiderivative, so it cannot be \(dv\). Choose \(u = \ln x\) and \(dv = \sqrt{x}\,dx\).
Differentiate \(u\), and integrate \(dv\). The corner of the orange rectangle is the product \(uv\).
\[du = \frac{1}{x}\,dx, \qquad v = \frac23 x^{3/2}\]The new integral multiplies \(v\) by \(du\). The factor \(\tfrac1x\) lowers the power of \(x\).
\[\int v\,du = \frac23\int x^{1/2}\,dx = \frac49 x^{3/2}\]The new integral is a power of \(x\), which you can integrate, so this split works. The formula gives \(uv\) minus the new integral.
\[\begin{aligned}\int \sqrt{x}\,\ln x\,dx &= \frac23 x^{3/2}\ln x \\ &\quad - \frac49 x^{3/2} + C\end{aligned}\]
Example 2: The power goes in u
Choose \(u\) and \(dv\) for \(\displaystyle\int x\sin 4x\,dx\). Then find the answer.
Both factors have antiderivatives, so the choice depends on which factor becomes simpler.
A tempting wrong move: Choose \(u = \sin 4x\) and \(dv = x\,dx\), because \(x\) is the simpler factor to integrate.
\[du = 4\cos 4x\,dx, \qquad v = \frac{x^2}{2}\]The new integral is \(\int 2x^2\cos 4x\,dx\), and the height of the red region is \(v = \tfrac{x^2}{2}\). The power of \(x\) went up from 1 to 2, so this integral is harder.
Reverse the roles. Choose \(u = x\), because its derivative is 1. Choose \(dv = \sin 4x\,dx\).
\[du = dx, \qquad v = -\frac{\cos 4x}{4}\]The new integral has no \(x\) in it.
\[\begin{aligned}\int v\,du &= -\frac14\int \cos 4x\,dx \\ &= -\frac{\sin 4x}{16}\end{aligned}\]The formula gives \(uv\) minus the new integral.
\[\begin{aligned}\int x\sin 4x\,dx &= -\frac{x\cos 4x}{4} \\ &\quad + \frac{\sin 4x}{16} + C\end{aligned}\]
Practice
Choose \(u\) and \(dv\) for \(\displaystyle\int \frac{\ln x}{\sqrt{x}}\,dx\), where \(x > 0\). Then find the answer.
Ask which factor has no basic antiderivative. The new integral should be a power of \(x\).
Solution
The function \(\ln x\) has no basic antiderivative, so it is \(u\). The other factor is \(dv\).
\[u = \ln x, \qquad dv = x^{-1/2}\,dx\]Differentiate \(u\), and integrate \(dv\).
\[du = \frac1x\,dx, \qquad v = 2\sqrt{x}\]The new integral is a power of \(x\), so the split works.
\[\int v\,du = \int 2x^{-1/2}\,dx = 4\sqrt{x}\]The formula gives \(uv\) minus the new integral.
\[\begin{aligned}\int \frac{\ln x}{\sqrt{x}}\,dx &= 2\sqrt{x}\ln x \\ &\quad - 4\sqrt{x} + C\end{aligned}\]
Questions
The figure shows the split \(u = x\) and \(dv = \cos x\,dx\) for \(\int_0^{\pi/2} x\cos x\,dx\). The curve plots \(v\) against \(u\).
Which region shows the new integral \(\int v\,du\)?
- The teal region under the curve
- The green region beside the curve
- The whole orange rectangle
Answer
The teal region under the curve
Which integral does the teal region under the curve show?
- \(\int \sin x\,dx\)
- \(\int x\sin x\,dx\)
- \(\int \cos x\,dx\)
Answer
\(\int \sin x\,dx\)
For \(\int x e^{5x}\,dx\), you choose \(u = x\) and \(dv = e^{5x}\,dx\). What is the new integral \(\int v\,du\)?
- \(\dfrac15\int e^{5x}\,dx\)
- \(\int 5e^{5x}\,dx\)
- \(\int \dfrac{5x^2}{2}e^{5x}\,dx\)
Answer
\(\dfrac15\int e^{5x}\,dx\)
Summary
You split a product into \(u\), the part you differentiate, and \(dv\), the part you integrate.
The split works when you can integrate \(dv\) and when \(u\) becomes simpler after you differentiate it.
Use it when the integrand is a product, such as \(x\) times an exponential or a logarithm.
This is one of four skills in topic 6.11 of Which & Why: AP Calculus.