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Difference between revisions of "Integration by parts"
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== Order ==
 
== Order ==
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Now, given an integrand, what should be <math>u</math> and what should be <math>dv</math>?  Since <math>u</math> will show up as <math>du</math> and <math>dv</math> as <math>v</math> in the integral on the RHS, u should be chosen such that it has an "easy" (or "easier") [[derivative]] and <math>dv</math> so that it has a easy [[antiderivative]].   
+
Now, given an integrand, what should be <math>u</math> and what should be <math>dv</math>?  Since <math>u</math> will show up as <math>du</math> and <math>dv</math> as <math>v</math> in the integral on the RHS, <math>u</math> should be chosen such that it has an "easy" (or "easier") [[derivative]] and <math>dv</math> so that it has a easy [[antiderivative]].   
  
 
A mnemonic for when to substitute <math>u</math> for what is LIATE:
 
A mnemonic for when to substitute <math>u</math> for what is LIATE:
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<math>\int xe^x\; dx=?</math>
 
<math>\int xe^x\; dx=?</math>
  
−
x has a pretty simple derivative, so let's say <math>u=x</math>.  Then <math>dv=e^x dx</math>, <math>du=dx</math>, and <math>v=\int dv=e^x</math>.  We have
+
<math>x</math> has a pretty simple derivative, so let's say <math>u=x</math>.  Then <math>dv=e^x dx</math>, <math>du=dx</math>, and <math>v=\int dv=e^x</math>.  We have
  
 
<math>\int xe^x\; dx=(x)(e^x)-\int (e^x)(dx)=xe^x-e^x + C=e^x(x-1) + C</math>, where <math>C</math> is the [[constant of integration]].  You can take the derivative to see that it is indeed our desired result.
 
<math>\int xe^x\; dx=(x)(e^x)-\int (e^x)(dx)=xe^x-e^x + C=e^x(x-1) + C</math>, where <math>C</math> is the [[constant of integration]].  You can take the derivative to see that it is indeed our desired result.

Latest revision as of 16:01, 11 March 2022

The purpose of integration by parts is to replace a difficult integral with an easier one. The formula is

$\int u\, dv=uv-\int v\,du$

Order

Now, given an integrand, what should be $u$ and what should be $dv$? Since $u$ will show up as $du$ and $dv$ as $v$ in the integral on the RHS, $u$ should be chosen such that it has an "easy" (or "easier") derivative and $dv$ so that it has a easy antiderivative.

A mnemonic for when to substitute $u$ for what is LIATE:

Logarithmic

Inverse trigonometric

Algebraic

Trigonometric

Exponential

If any two of these types of functions are in the function to be integrated, the type higher on the list should be substituted as u.

Examples

$\int xe^x\; dx=?$

$x$ has a pretty simple derivative, so let's say $u=x$. Then $dv=e^x dx$, $du=dx$, and $v=\int dv=e^x$. We have

$\int xe^x\; dx=(x)(e^x)-\int (e^x)(dx)=xe^x-e^x + C=e^x(x-1) + C$, where $C$ is the constant of integration. You can take the derivative to see that it is indeed our desired result.

See also

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