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Calculus

Understanding Integration by Parts in Calculus

10.29.2021 • 10 min read

Rachel McLean

Subject Matter Expert

Integration by parts is a handy method for integrating the product of two functions. In this article, we'll discuss the definition of this procedure and its formula, and then walk through how to integrate by parts and practice with some examples.

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  1. What Is Integration by Parts?

  2. Where Does Integration by Parts Come From?

  3. How to Integrate by Parts

  4. Examples of Integration by Parts

What Is Integration by Parts?

Integration by parts is the most useful integration technique for evaluating the integral of a product of functions. After separating a single function into a product of two functions, we can easily evaluate the function's integral by applying the integration by parts formula:

udv=uvvdu\int udv = uv - \int vdu

In this formula, dudu represents the derivative of uu, while vv represents the integral of dvdv. The integral of the product of uu and vv is then evaluated in terms of the integral of the product of dudu and vv. It's often much easier to solve the integral that results from this formula than it is to directly evaluate the original single function. Integration by parts gives us an efficient way to integrate functions that would otherwise be very difficult to integrate, and it's particularly helpful in tricky cases such as inverse trigonometric functions and logarithmic functions. 

Where Does Integration by Parts Come From?

You might be surprised to learn that the formula for integration by parts is derived directly from the familiar Product Rule for derivatives. Let's review the formula for the Product Rule, which states that the derivative of a product of two functions is equal to the first function times the derivative of the second, plus the second function times the derivative of the first:

d(uv)dx=udvdx+vdudx\frac{d(uv)}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

Multiplying both sides of the equation by the differential dxdx, we obtain:

d(uv)=udv+vdud(uv)= udv + vdu

Now, let's take the integral of both sides. Remember that the integral of a sum is simply equal to the sum of their integrals. Integrating both sides gives us the following:

d(uv)=udv+vdu\int d(uv) = \int udv + \int vdu
uv=udv+vduuv = \int udv + \int vdu

All that's left to do now is rearrange the equation! Let's subtract vdu\int vdu from both sides. Does the resulting equation look familiar?

udv=uvvdu\int udv = uv - \int vdu

In this way, we can use the Product Rule to obtain the formula for integration by parts. So, if you ever forget the formula for integration by parts during a test, you can rely on the Product Rule to help rediscover it!

How to Integrate by Parts

Let's outline some simple steps for how to integrate by parts:

  1. Choose uu and dvdv to separate the given function into a product of functions.

  2. Differentiate uu to find dudu, and integrate dvdv to find vv.

  3. Plug uu, vv, dudu, and dvdv into the integration by parts formula: udv=uvvdu\int udv = uv - \int vdu

  4. Solve, and simplify where needed.

Let's do one example together in greater detail. Suppose we want to evaluate xexdx\int xe^xdx. First, we separate the function into a product of two functions. To do this, we need to designate one expression as uu and one as dvdv. Let's choose u=xu = x. Since we've chosen uu, what's left over must determine dvdv. Then dv=exdxdv = e^xdx.

Our next step is to differentiate uu to find dudu. It's clear that dudx=dxdx=1\frac{du}{dx}=\frac{dx}{dx}=1. Then, multiplying both sides by the differential dxdx, we find that du=1dxdu = 1dx. Now, let's integrate dvdv to find vv. Since the integral of exe^x is exe^x itself, we have v=exv = e^x. Finally, we plug u=xu = x, v=exv = e^x, du=1dxdu = 1dx, and dv=exdxdv = e^xdx into the integration by parts formula. We have:

udv=uvvdu\int udv = uv - \int vdu

xexdx=xexex(1dx)\int xe^xdx = xe^x - \int e^x(1dx)

xexdx=xexex+C\int xe^xdx = xe^x - e^x + C

We've found that xexdx=xexex+C\int xe^xdx = xe^x - e^x + C. Remember to include the constant of integration in your final answer!

Now, the question we have to ask is: how do we know what to set equal to uu each time, and what to set equal to dvdv? This decision is where the difficulty lies for many students. If you have a strong intuition for what's easiest to integrate and what's easiest to differentiate, you can denote uu as the term you'd rather differentiate and dvdv as the term you'd rather integrate.

If you feel unsure about what would be easiest to differentiate and integrate, it can be helpful to rely upon the mnemonic “LIATE,” an acronym based on an ordered list of different function types. Usually, the higher a function sits on the list, the higher precedence it takes as uu. With this logic, LIATE can help guide your decision on which term to designate as uu:

  • Logarithmic functions: such as lnx\ln{x}, log3(x)\log_3(x), etc.

  • Inverse trigonometric functions: such as arcsin(x)\arcsin(x), arctan(x)\arctan(x), etc.

  • Algebraic functions: such as xx, x2x^2, 4x34x^3, etc.

  • Trigonometric functions: such as sin(x)\sin(x), cos(x)cos(x), etc.

  • Exponential functions: such as exe^x, 4x4^x, etc.

See if you recognize any of these five function types inside the integral that you are trying to evaluate, and then categorize the terms appropriately in the same order outlined by LIATE. Since functions that are higher on the list are prioritized as uu, simply designate uu as the term that appears highest on your list. Then, you can set dvdv equal to whatever is left over.

To remember the formula for integration by parts, it might be helpful to use another mnemonic device. One popular choice for remembering the right-hand side of the integration by parts formula is “ultraviolet voodoo,” where “ultraviolet” corresponds to uvuv and “voodoo” corresponds to vdu\int vdu. A well-liked mnemonic for Spanish speakers is, “Un día vi una vaca vestida de uniforme.” With this device, un ( uu ) día vi ( dvdv ) represents the left-side of the integration by parts formula, and una ( uu ) vaca ( vv ) vestida ( vv ) de uniforme (du)(du) represents the right-hand side. The integration by parts formula is included again below for your reference. Can you think of any other mnemonic devices?

udv=uvvdu\int udv = uv - \int vdu

More Examples of Integration by Parts

Let's work through a few more examples together.

1. Evaluate xsin(x)dx\int x\sin(x)dx.

  • Let u=xu=x and dv=sin(x)dxdv=\sin(x)dx.

  • Differentiating uu, we have du=1dxdu = 1dx.

  • Integrating dvdv, we find v=cos(x)v = - \cos(x).

  • Now, let's plug u=xu=x, v=cos(x)v = - \cos(x), du=1dxdu = 1dx, and dv=sin(x)dxdv=\sin(x)dx into the integration by parts formula.

  udv=uvvdu\int udv = uv - \int vdu

  xsin(x)dx=xcos(x)cos(x)dx\int x\sin(x)dx = -x \cos(x) - \int - \cos(x)dx

  xsin(x)dx=xcos(x)+sin(x)+C\int x\sin(x)dx = -x \cos(x) + \sin(x) + C

2. Evaluate xln(x)dx\int x \ln{(x)} dx.

  • Let u=ln(x)u = \ln{(x)} and dv=xdv = x.

  • Differentiating uu, we have du=1xdxdu = \frac{1}{x}dx.

  • Integrating dvdv, we have v=x22v = \frac{x^2}{2}.

  • Now, we plug u=ln(x)u = \ln{(x)}, v=x22v = \frac{x^2}{2}, du=1xdxdu = \frac{1}{x}dx, and dv=xdv = x into the integration by parts formula.

  udv=uvvdu\int udv = uv - \int vdu

  xln(x)=ln(x)x22x221xdx\int x\ln{(x)} = \ln{(x)} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{x}dx

  xln(x)=x22ln(x)12xdx\int x\ln{(x)} = \frac{x^2}{2} \ln{(x)} - \frac{1}{2} \int xdx

  xln(x)=x22ln(x)12x22\int x\ln{(x)} = \frac{x^2}{2} \ln{(x)} - \frac{1}{2} \cdot \frac{x^2}{2}

  xln(x)=x22ln(x)14x2+C\int x\ln{(x)} = \frac{x^2}{2} \ln{(x)} - \frac{1}{4}x^2 + C

3. Evaluate x2sin(x)dx\int x^2 \sin(x)dx.

  • Let u=x2u = x^2 and dv=sin(x)dxdv = \sin{(x)}dx.

  • Differentiating uu, we have du=2xdxdu = 2xdx.

  • Integrating dvdv, we have v=cos(x)v = - \cos{(x}).

  • Now, let's plug u=x2u = x^2, v=cos(x)v = - \cos{(x}), du=2xdxdu = 2xdx, and dv=sin(x)dxdv = \sin{(x)}dx into the integration by parts formula.

  udv=uvvdu\int udv = uv - \int vdu

  x2sin(x)dx=x2(cos(x))(cos(x))2xdx\int x^2 \sin(x)dx = x^2 (-\cos{(x)}) - \int (- \cos{(x)})2xdx

  x2sin(x)dx=x2cos(x)+2xcos(x)dx\int x^2 \sin(x)dx = -x^2\cos{(x)} + 2 \int x \cos{(x)}dx

Sometimes, you'll have to integrate by parts more than once in the same problem. In this case, we'll need to use it again to evaluate xcos(x)dx\int x \cos{(x)}dx.

  • Let u1=xu_1 = x and dv1=cos(x)dxdv_1 = \cos{(x)}dx.

  • Differentiating u1u_1, we have du1=1dxdu_1 = 1dx.

  • Integrating dv1dv_1, we have v1=sin(x)v_1 = \sin{(x)}.

  • Now, we plug u1=xu_1 = x, v1=sin(x)v_1 = \sin{(x)}, du1=1dxdu_1 = 1dx, and dv1=cos(x)dxdv_1 = \cos{(x)}dx into the integration by parts formula.

  udv=uvvdu\int udv = uv - \int vdu

  xcos(x)dx=xsin(x)sin(x)dx\int x \cos{(x)}dx = x \sin{(x)} - \int \sin{(x})dx

  xcos(x)dx=xsin(x)(cos(x))\int x \cos{(x)}dx = x \sin{(x)} -(- \cos{(x)})

  xcos(x)dx=xsin(x)+cos(x)+C\int x \cos{(x)}dx = x \sin{(x)} + \cos{(x)} + C

Now, we've found that xcos(x)dx=xsin(x)+cos(x)+C\int x \cos{(x)}dx = x \sin{(x)} + \cos{(x)} + C. We can use substitution to plug this value into our original formula to finish the problem. Note that since CC is arbitrary, we don't have to multiply it by 2 when distributing. Picking up where we left off, we have:

  x2sin(x)dx=x2cos(x)+2xcos(x)dx\int x^2 \sin(x)dx = -x^2\cos{(x)} + 2 \int x \cos{(x)}dx

  x2sin(x)dx=x2cos(x)+2(xsin(x)+cos(x)\int x^2 \sin(x)dx = -x^2\cos{(x)} + 2(x \sin{(x)} + \cos{(x)}

  x2sin(x)dx=x2cos(x)+2xsin(x)+2cos(x)+C\int x^2 \sin(x)dx = -x^2\cos{(x)} + 2x \sin{(x)} + 2 \cos{(x)} + C

As you can see, integration by parts is a valuable instrument in your mathematical toolbox.

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