Integration by Parts

This section covers:

Introduction to Integration by Parts

Integration by Parts is yet another integration trick that can be used when you have an integral that happens to be a product of algebraic, exponential, logarithm, or trigonometric functions.

The rule of thumb is to try to use U-Substitution, but if that fails, try Integration by Parts. Typically, Integration by Parts is used when two functions are multiplied together, with one that can be easily integrated, and one that can be easily differentiated.

It’s not always that easy though, as we’ll see below (but we’ll have some hints). And sometimes we have to use the procedure more than once!

Here is the formal definition:

Integration by Parts


For functions u and v that are functions of x with continuous derivatives, then:

                          \[\int {u\,dv = uv\,\, – } \int {v\,} du\]


Let’s do an example:

Solve the following integral using integration by parts:  \(\int {{x^3}} \,\ln x\,dx\).

Since we have a product of two functions, let’s “pick it apart” and use the integration by parts formula \(\int{{udv}}\,=uv-\int{{vdu\,}}\). First, decide what the u and dv parts should be. Since it’s must easier to get the derivative of \(\ln x\) than the integral, let \(u = \ln x,\,\,\,dv = {x^3}dx\). Then we have \(du=\frac{1}{x}\,dx\) and \(v=\int{{{{x}^{3}}dx}}=\frac{{{{x}^{4}}}}{4}\); we can throw away the “+ C” for now. Now plug into the formula:

\(\displaystyle \begin{align}\int{{u\,dv}}&=uv\,-\int{{v\,du}}\,=\ln x\cdot \frac{{{{x}^{4}}}}{4}\,-\int{{\left( {\frac{{{{x}^{4}}}}{4}\cdot \frac{1}{x}} \right)\,}}dx\\&=\frac{{{{x}^{4}}}}{4}\ln x\,\,-\int{{\frac{{{{x}^{3}}}}{4}\,}}dx\\&=\frac{{{{x}^{4}}}}{4}\ln x\,\,-\frac{{{{x}^{4}}}}{{16}}+C\end{align}\)

Guidelines for Integration by Parts using LIATE

The most difficult thing about Integration by Parts is 1) knowing if you should use it and 2) deciding how to pick apart the integral. As with any form of integration, if you get to the point where you’re not going anywhere, it’s not the form of integration to use.

Here are some hints:

It makes sense that we want to look at the integral and determine what’s easier to integrate (which should be the dv) and what’s easier to differentiate (to be the u). Also, it’s best to let dv be the most complicated part of the integrand, and it should fit a basic integration rule. And it’s best to let u be the part of the integrand with a derivative simpler than u. And always remember that the dv part always includes the dx of the original integral.

There’s another way that uses an acronym to help with Integration by Parts. It has to do with which function should be the u; whichever function comes first in the following list should be the u. It doesn’t work 100%, but it’s sure helpful:

LIATE Rule (use this order for picking u)

  1. Logarithmic Functions
  2. Inverse Trig Functions
  3. Algebraic Functions
  4. Trig Functions
  5. Exponential Functions

Integration by Parts Problems

Let’s do some problems; note that the answers are simplified in most cases.

Integration by Parts Problem


Find the indefinite integral:


\[\int {x{e^{ – x}}\,dx} \]



Since \(dv={{e}^{{-x}}}\,dx\) is the most complicated part, and \(u=x\) has a very simple derivative, pick \(u=x\). Also, if we use LIATE, an algebraic function comes before an exponential one for picking u, so again it makes sense to pick \(u=x\).


\(\begin{align}u &= x\\du &= 1dx = dx\end{align}\)         \(\begin{align}dv &= {e^{ – x}}\,dx\\v &= \int {dv = \int {{e^{ – x}}\,dx} } = – {e^{ – x}}\end{align}\)                     

\(\begin{align}\int {u\,dv} &= uv\,\, – \int {v\,du} \\\int {x{e^{ – x}}\,d} x &= x \cdot – {e^{ – x}} – \,\int { – {e^{ – x}}} dx = – {e^{ – x}}x – \left( {{e^{ – x}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= – x{e^{ – x}} – {e^{ – x}} + C = – {e^{ – x}}\left( {x + 1} \right) + C\end{align}\)

Find the indefinite integral:


\[\int {x\sqrt {x – 4} \,} dx\]



Since the second part is more complicated, let’s try to set that to dv, and set the x to u:


\(\begin{align}u &= x\\du& = 1dx = dx\end{align}\)       \(\displaystyle \begin{align}dv&=\sqrt{{x-4}}\,dx={{\left( {x-4} \right)}^{{\frac{1}{2}}}}dx\\v&=\int{{dv=\int{{{{{\left( {x-4} \right)}}^{{\frac{1}{2}}}}dx}}}}=\frac{2}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\end{align}\)  

\[\begin{align}\int {u\,dv} &= uv\,\, – \int {v\,du} \\\int {x\sqrt {x – 4} \,dx} &= \,x \cdot \frac{2}{3}{\left( {x – 4} \right)^{\frac{3}{2}}} – \int {\frac{2}{3}\,{{\left( {x – 4} \right)}^{\frac{3}{2}}}dx = \frac{{2x}}{3}{{\left( {x – 4} \right)}^{\frac{3}{2}}} – \frac{2}{3} \cdot \frac{2}{5}\,{{\left( {x – 4} \right)}^{\frac{5}{2}}} + C} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{{2x}}{3}{\left( {x – 4} \right)^{\frac{3}{2}}} – \frac{4}{{15}}\,{\left( {x – 4} \right)^{\frac{5}{2}}} + C = \frac{2}{{15}}{\left( {x – 4} \right)^{\frac{3}{2}}}\left[ {5x – 2\left( {x – 4} \right)} \right] + C\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= \frac{2}{{15}}{\left( {x – 4} \right)^{\frac{3}{2}}}\left( {3x + 8} \right) + C\end{align}\]




Find the indefinite integral: 


\[\int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} \,dx\]


Let’s try to make the \(\frac{1}{{{{{\left( {x+1} \right)}}^{2}}}}dx\) the dv part, since we can easily take the integral of this. (Remember if the choices for u and dv don’t work out the first time you try it, you can try to rearrange differently and try again.)


\(\begin{align}u &= x{e^x}\\du &= \left( {x \cdot {e^x} + {e^x} \cdot 1} \right)dx\\ &= {e^x}\left( {x + 1} \right)dx\end{align}\)         \(\begin{align}dv &= \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \,dx = {\left( {x + 1} \right)^{ – 2}}dx\\v &= \int {dv = \int {{{\left( {x + 1} \right)}^{ – 2}}} dx} = – {\left( {x + 1} \right)^{ – 1}}\end{align}\) 

\int {u\,dv} &= uv\,\, – \int {v\,du} \\
\int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} \,dx\, &= \left[ {x{e^x} \cdot – {{\left( {x + 1} \right)}^{ – 1}}} \right] – \int {\left[ { – {{\left( {x + 1} \right)}^{ – 1}} \cdot {e^x}\left( {x + 1} \right)} \right]dx} \\
\,\,\,\, &= \frac{{ – x{e^x}}}{{x + 1}} – \int {\left[ { – \frac{{{e^x}\left( {x + 1} \right)}}{{x + 1}}} \right]dx} = \frac{{ – x{e^x}}}{{x + 1}} + \int {{e^x}dx = } \frac{{ – x{e^x}}}{{x + 1}} + {e^x} + C \\
\,\,\,\, &= \frac{{ – x{e^x}}}{{x + 1}} + \frac{{{e^x}\left( {x + 1} \right)}}{{x + 1}} + C = \frac{{ – x{e^x} + x{e^x} + {e^x}}}{{x + 1}}  + C = \frac{{{e^x}}}{{x + 1}} + C \\



Sometimes, we have to use Integration by Parts twice. It turns out that we had to perform the Integration by Parts twice, since we had an  before the . If we had an , we would have had to do it three times. We’ll show an easier (tabular) way to handle these situations below.

Sometimes, we have to use Integration by Parts twice, as in the following example, where we have an \({{x}^{2}}\) before the \(\sin x\). If we had an \({{x}^{3}}\), we would have had to do it three times. We’ll show an easier (tabular) way to handle these situations below.

Integration Problem






Find the indefinite integral:

\[\int {{x^2}\sin x\,} dx\]




Since trig functions usually follow algebraic functions in determining u, let’s pick  and :


\(\begin{align}u &= {x^2}\\du &= 2x\,dx\end{align}\)         \(\begin{align}dv &= \sin x\,dx\\v &= \int {dv = \int {\sin x\,dx} } = – \cos x\end{align}\)

\[\begin{align}\int {u\,dv} &= \int {uv} \,\,\, – \,\int {v\,} du\\\int {{x^2}\sin x\,dx} &= {x^2} \cdot – \cos x\,\, – \,\int { – \cos x \cdot 2x\,} dx\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= {\color{red}{- {x^2}\cos x} } +{\color{blue}{\,\int {2x\cos xdx}}}\,\,\end{align}\]


Integration by parts second time:

\(\begin{align}u &= {\color{blue}{2x}}\\du &= 2\,dx\end{align}\)     \(\begin{align}dv &= {\color{blue}{\cos x\,dx}}\\v &= \int {dv = \int {\cos x\,dx} } = \sin x\end{align}\)


\int {u\,dv} &= \,uv\,\, – \int {vdu\,} \\
{\color{blue}{ \int {2x\cos xdx}}} &= 2x \cdot \sin x\,\, – \int {\sin x \cdot 2dx} = 2x\sin x – 2( – \cos x) + C\, \\
&= {\color{blue}{\,2x\sin x + 2\cos x + C}} \\
\end{align} \]


Now put it all together:

\[\int {{x^2}\sin x\,} dx = {\color{red}{- {x^2}\cos x} } + {\color{blue}{\,2x\sin x + 2\cos x + C}}\, = \left( {2 – {x^2}} \right)\cos x + 2x\sin x + C\]



Tabular Method for Integration by Parts

Instead of performing Integration of Parts over and over again (like the problem above), there is a much easier way to solve using a table. These problems typically have x raised to a power; we have to get that power down to 0 in order to solve.

Make the u part, the variable raised to the power, and the vdv part what we call the “indestructible” part: the part that doesn’t go down to 0 when you keep taking it’s derivative or integral.

Start with the u part of the equation (usually the variable raised to the power), and in the second column, keep taking derivatives until you reach 0. In the next column, start with the “dv” part, and keep taking integrals for every row.

Then multiply the second and third columns (ignoring the first term of the third column) and use alternate signs, as shown in the first column (start with a +). Here is an example:

Find \(\int{{{{x}^{3}}}}\cos \left( {2x} \right)dx\) using Integration by Parts tabular method: \(u={{x}^{3}};\,\,\,\,v\,dv=\cos \left( {2x} \right)dx\).

Tabular Method Integration By Parts

To get the integral, multiply across the arrows of columns two and three, using the signs from column one:

\int {{x^3}} \cos \left( {2x} \right)dx &= {x^3} \cdot \frac{1}{2}\sin \left( {2x} \right) – 3{x^2} \cdot – \frac{1}{4}\cos \left( {2x} \right) + 6x \cdot – \frac{1}{8}\sin \left( {2x} \right) – 6 \cdot \frac{1}{{16}}\cos \left( {2x} \right) + C \\
&= \frac{1}{2}{x^3}\sin \left( {2x} \right) + \frac{3}{4}{x^2}\cos \left( {2x} \right) – \frac{3}{4}x\sin \left( {2x} \right) – \frac{3}{8}\cos \left( {2x} \right) + C \\
&= \frac{1}{8}\left( {4{x^3}\sin \left( {2x} \right) + 6{x^2}\cos \left( {2x} \right) – 6x\sin \left( {2x} \right) – 3\cos \left( {2x} \right)} \right) + C \\

Let’s try one more:

Find \(\int{{{{x}^{4}}}}{{e}^{{-2x}}}dx\) using Integration by Parts tabular method: \(u={{x}^{4}};\,\,\,\,v\,dv={{e}^{{-2x}}}dx\).

Tabular Method Integration By Parts 2

To get the integral, multiply across the arrows of columns two and three, using the signs from column one:

\int {{x^4}} {e^{ – 2x}}dx &= {x^4} \cdot – \frac{1}{2}{e^{ – 2x}} – 4{x^3} \cdot \frac{1}{4}{e^{ – 2x}} + 12{x^2} \cdot – \frac{1}{8}{e^{ – 2x}} – 24x \cdot \frac{1}{{16}}{e^{ – 2x}} + 24 \cdot – \frac{1}{{32}}{e^{ – 2x}} + C \\
&= – \frac{1}{2}{x^4}{e^{ – 2x}} – {x^3}{e^{ – 2x}} – \frac{3}{2}{x^2}{e^{ – 2x}} – \frac{3}{2}x{e^{ – 2x}} + – \frac{3}{4}{e^{ – 2x}} + C \\
&= – \frac{1}{4}{e^{ – 2x}}\left( {2{x^4} + 4{x^3} + 6{x^2} + 6x + 3} \right) + C \\

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On to Integration by Partial Fractions – you’re ready!