If you need some help with integration by parts, then you are in the right place. This review article will give you a simple guide, and link integration by parts to multivariable calculus.

Most people find differentiation easier than integration, because differentiation has familiar and well-defined rules. Integrating functions can be more complicated and requires more experience, particularly because there are several different ways to integrate. Although integrating basic functions can be simple, things can get complicated when integrating products of functions. One of the most common ways to integrate products of functions is the “integrating by parts” method. Note that integration by parts is not used when the integral in question is a product of the integrals of the individual functions.

There are a few things that we need to define before we see how to integrate by parts. If *u *and *v* are functions of *x*, then we have:

\frac { d\left( uv \right) }{ dx } =u\left( x \right) \frac { dv\left( x \right) }{ dx } +v\left( x \right) \frac { du\left( x \right) }{ dx }

The above equation is known as the product rule. It says that the derivative of a product of two functions is the sum of the first function times the derivative of the second, plus the second function times the derivative of the first function. Rearranging this equation gives us:

u\left( x \right) \frac { dv\left( x \right) }{ dx } =v\left( x \right) \frac { du\left( x \right) }{ dx } -\frac { d\left( uv \right) }{ dx }

Integrating the above equation with respect to *x* gives us the following:

\int { udv } =uv-\int { vdu }

The above equation is the integration by parts formula. This allows us to turn complicated integrals into simpler integrals. The left-hand side of the equation is called the integrand. Unlike the product rule, from which the integration by parts formula is derived, we do not identify *u* and *v* but *u* and *dv*. Don’t be scared that it’s made up of the derivative of the function *v*. All you need to do is choose the correct functions for *u* and *dv*.

So now that we have the formula, what next? The first thing to do is identify the two basic functions of the product and substitute them for *u* and *dv* in the integrand. There are many methods to help us choose which function to use for *u. *

A good rule of thumb for choosing *u *is to remember the abbreviation LIATE; this stands for Logarithmic, Inverse circular, Algebraic, Trigonometric (circular), and Exponential. Your first choice should be logarithmic, followed by inverse circular, and so on. The function for *u* should be chosen such that *du* is of a simpler form than *u*. If there is a power function, then it is a good idea to use this as *u*, as integrating this function will reduce the power by one, resulting in a simpler function on the right-hand side of the integration by parts formula. Obviously, choosing *u* means that the remaining function is assigned to *dv*.

Once *u* and *dv* have been identified, you need to calculate *v* and *du*. To do this, you differentiate *u* to get *du* and integrate *dv* to get *v*. Then, *u* and *dv* are substituted into the right-hand side of the integration by parts equation. The right-hand side presents the result of the integral. If *u* and *v* have been chosen correctly, the integral on the right-hand side should just end up being a simple function, or simpler than the original function.

Now, let’s try some integration by parts examples together.

**Example 1:**

\int { x\sin { x } } dx

Following the rule that *du* should be of a simpler form than *u*, we chose *u* = *x.* Hence, *dv* = *sinxdx. *

Integrating *u*, we have *du = dx*, and differentiating *dv*, we have:

v=\int { \sin { x } dx } =-\cos { x }

Putting these into the integration by parts equation, we have:

\int { udv } =uv-\int { vdu }

\int { x\sin { x } dx } =(x)\left( -\cos { x } \right) -\int { (-\cos { x )} }

=-x\cos { x } +\sin { x } +c

**Example 2:**

\int { { x }^{ 2 }{ e }^{ -x } } dx

Using the LIATE rule, we can let *u = x ^{2} *and

\int { udv } =uv-\int { vdu }

\int { { x }^{ 2 }{ e }^{ -x } } dx=({ x }^{ 2 })({ -e }^{ -x })-\int { ({ -e }^{ -x })(2x)dx }

={ -x }^{ 2 }{ e }^{ -x }+2\int { { xe }^{ -x }dx }

The integral above is not a simple function; we need to perform integration by parts again. Let *u = x* and hence, *dv = e ^{-x }dx.* Therefore, we have

\int { udv } =uv-\int { vdu }

\int { { xe }^{ -x }dx } =(x)({ -e }^{ -x })-\int { ({ -e }^{ -x })dx }

={ -xe }^{ -x }+\int { ({ e }^{ -x })dx }

={ -xe }^{ -x }-{ e }^{ -x }

Putting both parts together, we have:

\int { { x }^{ 2 }{ e }^{ -x } } dx={ -x }^{ 2 }{ e }^{ -x }+2({ -xe }^{ -x }-{ e }^{ -x })

={ -e }^{ -x }({ x }^{ 2 }+2x+2)+c

**Example 3:**

\int { { x }^{ 2 }\ln { x } } dx

Following LIATE, we shall choose *lnx* for *u* and hence, *dv = x ^{2} dx*. This means that we have:

du=\frac { dx }{ x }

And

v=\frac { 1 }{ 3 } { x }^{ 3 }

Substituting this into the integration by parts formula, we have:

\int { udv } =uv-\int { vdu }

\int { { x }^{ 2 }\ln { x } dx } =(\ln { x } )(\frac { 1 }{ 3 } { x }^{ 3 })-\int { (\frac { 1 }{ 3 } { x }^{ 3 }) } (\frac { dx }{ x } )

=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } -\frac { 1 }{ 3 } \int { ({ x }^{ 2 })dx }

=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } -(\frac { 1 }{ 3 }) (\frac {{ x }^{ 3 }}{3})+c

=\frac { 1 }{ 3 } { x }^{ 3 }\ln { x } - \frac {{ x }^{ 3 }}{9}+c

Things can get a bit more complicated when we start dealing with multivariable functions (functions that depend on more than one variable).

Let *u(x _{1}, …, x_{n})* and

Let us denote \Omega as an open, bounded subset of *R*^{n}, with a boundary \Gamma . If we assume that *u* and *v* are continuously differentiable over \Omega , then the standard integration by parts formula can be adapted to multivariate functions as follows:

\int _{ \Omega } \frac { \partial u }{ \partial { x }_{ i } } vd\Omega =\int _{ \Gamma } uv{ v }_{ i }d\Gamma -\int _{ \Omega } \frac { \partial v }{ \partial { x }_{ i } } vd\Omega

Here, v\prime is the outward unit normal to the boundary \Gamma , and *v _{i}* is its

The above equation can be generalized by substituting *v _{i}* with

\int _{ \Omega } \nabla u.vd\Omega =\int _{ \Gamma } (uv).v\prime d\Gamma -\int _{ \Omega }u\nabla .vd\Omega

Here, *v* is a vector function with components *v _{1}, …, v_{n}*. The above equation leads us to an important theorem when

\int _{ \Omega } \nabla .vd\Omega =\int _{ \Gamma }v.v\prime d\Gamma

Likewise, we also obtain Green’s formula:

\int _{ \Omega } \nabla u.\nabla vd\Omega =\int _{ \Gamma } u\nabla v.v\prime d\Gamma -\int _{ \Omega } u\nabla vd\Omega

The divergence theorem is used to relate surface integrals with triple integrals. For example, it relates a volume integral over volume \Omega (the left-hand side of the equation) to a surface integral over the surface enclosing the volume (the right-hand side). Similarly, Green’s formula can be used to relate line integrals to area integrals.

We hope that you have found this guide to integration by parts useful. After reading this review article, you will have seen where the integration by parts formula comes from, why it is used, and, most importantly, you should be able to tackle even the most difficult integration by parts problems with the tips given above. You will also have read how the integration by parts formula is adapted for multivariable functions so that it can be used for multidimensional problems.

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