Introduction to Integration by Parts | Tabular Method for Integration by Parts |

Guidelines for Integration by Parts using LIATE | More Practice |

Integration by Parts Problems |

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 Integration, 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:

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

Here is an example:

Integration by Parts Problem | Solution |

Solve the following integral using integration by parts:
\(\displaystyle \int {{x^3}} \,\ln x\,dx\) |
Since we have a product of two functions, let’s “pick it apart” and use the integration by parts formula \(\displaystyle \int{{udv}}\,=uv-\int{{vdu\,}}\).
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 \(\displaystyle du=\frac{1}{x}\,dx\) and \(\displaystyle v=\int{{{{x}^{3}}dx}}=\frac{{{{x}^{4}}}}{4}\); we can throw away the “\(+\,C\)” for now. Now plug into the formula: \(\displaystyle \begin{align}\int{{u\,dv}}&=uv\,-\int{{v\,du}}\,=\ln x\cdot \frac{{{{x}^{4}}}}{4}\,-\int{{\left( {\frac{{{{x}^{4}}}}{4}\cdot \frac{1}{x}} \right)\,}}dx\\&=\frac{{{{x}^{4}}}}{4}\ln x\,\,-\int{{\frac{{{{x}^{3}}}}{4}\,}}dx=\frac{{{{x}^{4}}}}{4}\ln x\,\,-\frac{{{{x}^{4}}}}{{16}}+C=\frac{{{{x}^{4}}}}{{16}}\left( {4\ln x\,\,-1} \right)+C\end{align}\) Pretty cool! |

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. And if it doesn’t work the first time, switch the integral parts and try again! Here are some hints that might help:

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 \(u\) itself, if you can. And always remember that the \(dv\) part always includes the “\(dx\)” of the original integral.

There’s a trick that uses an acronym to help. It has to do with which function should be the \(u\). It doesn’t work 100%, but it’s sure helpful; the acronym is LIATE; use this order for picking \(u\). (ILATE works too, since you usually don’t have an inverse trig function and logarithmic function in same problem). Pick the \(u\) in this order:

- Logarithmic Functions, such as \(\log \left( x \right),\ln \left( x \right)\)
- Inverse Trigonometric Functions, such as \({{\sin }^{{-1}}}\left( x \right)\)
- Algebraic Functions, such as \(x,{{x}^{4}}\)
- Trigonometric Functions, such as \(\sin \left( x \right)\)
- Exponential Functions, such as \({{2}^{x}},{{e}^{x}}\)

We’ll show an example below.

Integration by Parts Problems

Here are some problems; note that the answers are simplified in most cases.

Integration by Parts Problem | Solution |

Find the indefinite integral:
\[\int {x{e^{ – x}}\,dx} \] |
Since \(\displaystyle dv={{e}^{{-x}}}\,dx\) 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 \(u=x\).
\(\begin{align}u &= x\\du &= 1dx = dx\end{align}\) \(\begin{align}dv &= {e^{ – x}}\,dx\\v &= \int {dv = \int {{e^{ – x}}\,dx} } = – {e^{ – x}}\end{align}\) \(\begin{align}\int{{u\,dv}}&=uv-\int{{v\,du}}\\\int{{x{{e}^{{-x}}}\,d}}x&=x\cdot \left( {-{{e}^{{-x}}}} \right)-\,\int{{-{{e}^{{-x}}}}}dx=-{{e}^{{-x}}}x-\left( {{{e}^{{-x}}}} \right)\\&=-x{{e}^{{-x}}}-{{e}^{{-x}}}+C=-{{e}^{{-x}}}\left( {x+1} \right)+C\end{align}\) |

Find the indefinite integral:
\[\int {x\sqrt {x – 4} \,} dx\] |
Since the second part is more complicated, try to set that to \(dv\), and set the \(x\) to \(u\):
\(\begin{align}u &= x\\du& = 1dx = dx\end{align}\) \(\displaystyle \begin{align}dv&=\sqrt{{x-4}}\,dx={{\left( {x-4} \right)}^{{\frac{1}{2}}}}dx\\v&=\int{{dv=\int{{{{{\left( {x-4} \right)}}^{{\frac{1}{2}}}}dx}}}}=\frac{2}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\end{align}\) \(\displaystyle \begin{align}\int{{u\,dv}}&=uv\,\,-\int{{v\,du}}\\\int{{x\sqrt{{x-4}}\,dx}}&=\,x\cdot \frac{2}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}-\int{{\frac{2}{3}\,{{{\left( {x-4} \right)}}^{{\frac{3}{2}}}}dx=\frac{{2x}}{3}{{{\left( {x-4} \right)}}^{{\frac{3}{2}}}}-\frac{2}{3}\cdot \frac{2}{5}\,{{{\left( {x-4} \right)}}^{{\frac{5}{2}}}}+C}}\\&=\frac{{2x}}{3}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}-\frac{4}{{15}}\,{{\left( {x-4} \right)}^{{\frac{5}{2}}}}+C=\frac{2}{{15}}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\left[ {5x-2\left( {x-4} \right)} \right]+C=\frac{2}{{15}}{{\left( {x-4} \right)}^{{\frac{3}{2}}}}\left( {3x+8} \right)+C\end{align}\) Note the factoring we learned about here in the Advanced Factoring section. |

Find the indefinite integral: \[\int {\frac{{x{e^x}}}{{{{\left( {x + 1} \right)}^2}}}} \,dx\] |
Try to make the \(\displaystyle \frac{1}{{{{{\left( {x+1} \right)}}^{2}}}}dx\) 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.)
\(\begin{align}u &= x{e^x}\\du &= \left( {x \cdot {e^x} + {e^x} \cdot 1} \right)dx\\ &= {e^x}\left( {x + 1} \right)dx\end{align}\) \(\begin{align}dv &= \int {\frac{1}{{{{\left( {x + 1} \right)}^2}}}} \,dx = {\left( {x + 1} \right)^{ – 2}}\,dx\\v &= \int {dv = \int {{{\left( {x + 1} \right)}^{ – 2}}} \,dx} = – {\left( {x + 1} \right)^{ – 1}}\end{align}\) \(\displaystyle \begin{align}\int{{u\,dv}}&=uv\,-\int{{v\,du}}\\\int{{\frac{{x{{e}^{x}}}}{{{{{\left( {x+1} \right)}}^{2}}}}}}\,dx\,&=\left[ {x{{e}^{x}}\cdot -{{{\left( {x+1} \right)}}^{{-1}}}} \right]-\int{{\left[ {-{{{\left( {x+1} \right)}}^{{-1}}}\cdot {{e}^{x}}\left( {x+1} \right)} \right]dx}}\\\,&=\frac{{-x{{e}^{x}}}}{{x+1}}-\int{{\left[ {-\frac{{{{e}^{x}}\left( {x+1} \right)}}{{x+1}}} \right]dx}}=\frac{{-x{{e}^{x}}}}{{x+1}}+\int{{{{e}^{x}}dx=}}\frac{{-x{{e}^{x}}}}{{x+1}}+{{e}^{x}}+C\\\,&=\frac{{-x{{e}^{x}}}}{{x+1}}+\frac{{{{e}^{x}}\left( {x+1} \right)}}{{x+1}}+C=\frac{{-x{{e}^{x}}+x{{e}^{x}}+{{e}^{x}}}}{{x+1}}=\frac{{{{e}^{x}}}}{{x+1}}+C\end{align}\) |

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 | Solution |

Find the indefinite integral: \[\int {{x^2}\sin x\,} dx\] |
Since trig functions usually follow algebraic functions in determining \(u\) (LIATE), pick \(u={{x}^{2}}\) and \(dv=\sin x\,dx\):
\(\begin{align}u &= {x^2}\\du &= 2x\,dx\end{align}\) \(\begin{align}dv &= \sin x\,dx\\v &= \int {dv = \int {\sin x\,dx} } = – \cos x\end{align}\) \(\displaystyle \begin{align}\int{{u\,dv}}&=\int{{uv}}\,\,\,-\,\int{{v\,}}du\\\int{{{{x}^{2}}\sin x\,dx}}&={{x}^{2}}\cdot \left( {-\cos x} \right)-\,\int{{-\cos x\cdot \left( {2x} \right)\,}}dx\\&= {\color{red}{- {x^2}\cos x} } +{\color{blue}{\,\int {2x\cos xdx}}}\,\,\end{align}\) Integration by parts second time: \(\begin{align}u &= {\color{blue}{2x}}\\du &= 2\,dx\end{align}\) \(\begin{align}dv &= {\color{blue}{\cos x\,dx}}\\v &= \int {dv = \int {\cos x\,dx} } = \sin x\end{align}\) \[\begin{align} Now put it all together: \(\displaystyle \int {{x^2}\sin x\,} dx = {\color{red}{- {x^2}\cos x} } + {\color{blue}{\,2x\sin x + 2\cos x + C}}\, = \left( {2 – {x^2}} \right)\cos x + 2x\sin x + C\) |

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. The \(u\)-part is typically the variable raised to the power, and the \(v\,dv\)-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 (for example, \({{e}^{x}}\) or \(\sin x\)).

Start with the \(u\)-part of the equation (again, 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 \(\displaystyle \int{{{{x}^{3}}}}\cos \left( {2x} \right)\,dx\) using Integration by Parts tabular method: \(u={{x}^{3}};\,\,\,\,v\,dv=\cos \left( {2x} \right)\,dx\).

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

\(\begin{align}\int{{{{x}^{3}}}}\cos \left( {2x} \right)dx&={{x}^{3}}\cdot \left[ {\frac{1}{2}\sin \left( {2x} \right)} \right]-3{{x}^{2}}\cdot \left[ {-\frac{1}{4}\cos \left( {2x} \right)} \right]+6x\cdot \left[ {-\frac{1}{8}\sin \left( {2x} \right)} \right]-6\cdot \left[ {\frac{1}{{16}}\cos \left( {2x} \right)} \right]+C\\&=\frac{1}{2}{{x}^{3}}\sin \left( {2x} \right)+\frac{3}{4}{{x}^{2}}\cos \left( {2x} \right)-\frac{3}{4}x\sin \left( {2x} \right)-\frac{3}{8}\cos \left( {2x} \right)+C\\&=\frac{1}{4}x\left( {2{{x}^{2}}-3} \right)\sin \left( {2x} \right)+\frac{3}{8}\left( {2{{x}^{2}}-1} \right)\cos \left( {2x} \right)+C\end{align}\)

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

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

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

Understand these problems, and practice, practice, practice!

On to Integration by Partial Fractions – you are ready!

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