Integration By Parts Examples And Solutions

You're staring at an integral. It's a product of two functions — maybe a polynomial times an exponential, or a logarithm times something algebraic. You try u-substitution. Here's the thing — nothing cancels. On top of that, you try to expand. The algebra explodes. And you're wondering: is there actually a way through this?

There is. It's called integration by parts. And if you've taken calculus, you've seen the formula. But seeing the formula and knowing when to use it — let alone how to pick u and dv without guessing — are completely different things.

What Is Integration by Parts

Integration by parts is the integration counterpart of the product rule for differentiation. That's the shortest way to say it. The product rule says:

d/dx [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)

Integrate both sides with respect to x:

∫ d/dx [u(x)v(x)] dx = ∫ u'(x)v(x) dx + ∫ u(x)v'(x) dx

The left side is just u(x)v(x). Rearrange:

∫ u(x)v'(x) dx = u(x)v(x) - ∫ u'(x)v(x) dx

That's it. In the standard notation:

∫ u dv = uv - ∫ v du

You're trading one integral for another. The hope — the whole point — is that the new integral is simpler than the one you started with.

The LIATE Heuristic (And Why It's Not a Law)

Most textbooks give you LIATE to choose u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Pick u from the leftmost category that appears. dv gets the rest.

It works a surprising amount of the time. But it's a heuristic, not a theorem. There are integrals where LIATE leads you in circles or makes things worse. The real skill is recognizing structure — not following a mnemonic blindly.

Why It Matters / Why People Care

Because a huge chunk of integrals you'll meet in calculus II, differential equations, physics, and engineering are products. And most of them don't yield to u-substitution.

Fourier series? On the flip side, probability density functions involving polynomials times exponentials? Integration by parts. Now, integration by parts. Integration by parts. So the gamma function? In real terms, laplace transforms? You guessed it.

If you can't do integration by parts fluently, you hit a wall in every applied math course after this one. In real terms, it's not a trick. It's a fundamental tool.

And here's the thing most students miss: you often need to apply it more than once*. Sometimes three times. Sometimes you get back to the original integral and solve algebraically. That last one feels like magic the first time you see it — but it's just algebra.

How It Works (or How to Do It)

Step 1: Identify the Product

Look at the integrand. Consider this: is it a product of two functions? Can you make* it a product? On the flip side, (Hint: ln(x) is ln(x) * 1. arctan(x) is arctan(x) * 1. This comes up more than you'd think.

Step 2: Choose u and dv

This is where the art lives. You want:

• du to be simpler than u (or at least no worse)
• v to be easy to find from dv
• The new integral ∫ v du to be doable

Let's walk through the classic categories.

Example 1: Polynomial Times Exponential

∫ x e^x dx

LIATE says algebraic before exponential. So u = x, dv = e^x dx.

Then du = dx, v = e^x.

∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C = e^x(x - 1) + C

Check by differentiating: derivative of e^x(x - 1) is e^x(x - 1) + e^x = x e^x. Works.

Example 2: Polynomial Times Trig

∫ x^2 sin(x) dx

u = x^2, dv = sin(x) dx du = 2x dx, v = -cos(x)

∫ x^2 sin(x) dx = -x^2 cos(x) - ∫ -cos(x) * 2x dx = -x^2 cos(x) + 2∫ x cos(x) dx

Now you have ∫ x cos(x) dx. Do parts again.

u = x, dv = cos(x) dx du = dx, v = sin(x)

∫ x cos(x) dx = x sin(x) - ∫ sin(x) dx = x sin(x) + cos(x)

Plug it back:

∫ x^2 sin(x) dx = -x^2 cos(x) + 2[x sin(x) + cos(x)] + C = -x^2 cos(x) + 2x sin(x) + 2 cos(x) + C

Notice the pattern: each application reduces the polynomial power by one. That's why it terminates.

Example 3: Logarithmic Function

∫ ln(x) dx

This doesn't look* like a product. But ln(x) = ln(x) * 1.

u = ln(x), dv = dx du = (1/x) dx, v = x

∫ ln(x) dx = x ln(x) - ∫ x * (1/x) dx = x ln(x) - ∫ 1 dx = x ln(x) - x + C

This exact pattern — logarithmic times 1 — shows up constantly. Memorize the result: ∫ ln(x) dx = x ln(x) - x + C. Same for ∫ log_a(x) dx, just adjust the constant.

Example 4: Inverse Trig Function

∫ arctan(x) dx

Again, write as arctan(x) * 1.

u = arctan(x), dv = dx du = 1/(1+x^2) dx, v = x

∫ arctan(x) dx = x arctan(x) - ∫ x/(1+x^2) dx

The remaining integral is a u-sub: let w = 1+x^2, dw = 2x dx.

∫ x/(1+x^2) dx = (1/2) ln(1+x^2) + C

So ∫ arctan(x) dx = x arctan(x) - (1/2) ln(1+x^2) + C

Example 5: The Boomerang Integral (Exponential Times Trig)

∫ e^x sin(x) dx

This is the one that makes students panic. Neither function gets simpler when differentiated. But watch.

First pass: u = sin(x), dv = e^x dx du = cos(x) dx, v = e^x

∫ e^x sin(x) dx = e^x sin(x) - ∫ e^x cos(x) dx

Second pass on the new integral: u = cos(x), dv = e^x dx du = -sin(x) dx, v = e^x

∫

Finishing the Boomerang Integral

Let’s complete the calculation that was left hanging.
Define

[ I=\int e^{x}\sin x,dx . ]

Apply integration by parts with (u=\sin x) and (dv=e^{x}dx).
Then (du=\cos x,dx) and (v=e^{x}), giving

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[ I=e^{x}\sin x-\int e^{x}\cos x,dx . ]

Now tackle the new integral (\displaystyle J=\int e^{x}\cos x,dx) in the same way, choosing (u=\cos x) and (dv=e^{x}dx).
Thus (du=-\sin x,dx) and (v=e^{x}), so

[ J=e^{x}\cos x+\int e^{x}\sin x,dx = e^{x}\cos x+I . ]

Substituting this expression for (J) back into the formula for (I) yields

[ I=e^{x}\sin x-\bigl(e^{x}\cos x+I\bigr) =e^{x}\sin x-e^{x}\cos x-I . ]

Collect the terms containing (I) on one side:

[ I+I = e^{x}\bigl(\sin x-\cos x\bigr) . ]

Hence

[ 2I = e^{x}\bigl(\sin x-\cos x\bigr)+C, \qquad\text{so}\qquad I = \frac{e^{x}}{2}\bigl(\sin x-\cos x\bigr)+C . ]

A quick differentiation confirms the result:

[ \frac{d}{dx}!\left[\frac{e^{x}}{2}\bigl(\sin x-\cos x\bigr)\right] = e^{x}\sin x . ]

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A Quick

A Quick Guide to Choosing (u) and (dv)

Once you stare at an integral that looks like a product, the first question is which part to differentiate and which part to integrate. A handy mental shortcut is the LIATE rule:

• Logarithmic ((\ln x), (\log_a x))
• Inverse trigonometric ((\arcsin x), (\arctan x), …)
• Algebraic ((x^n), polynomials)
• Trigonometric ((\sin x), (\cos x), …)
• Exponential ((e^{x}), (a^{x}))

Pick the function that appears earlier in the list for (u); the remainder becomes (dv). This heuristic works for most textbook problems and helps you avoid the “boomerang” trap where both pieces stay stubborn.

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Tabular Integration: A Shortcut for Repeated Parts

If you find yourself applying integration by parts multiple times with the same two families of functions (e.g., a polynomial times an exponential or a trig function), you can streamline the process with a tabular approach.

Derivative of (u)	Integral of (dv)
(u)	(dv)
(u')	(\int dv)
(u'')	(\int!!\int dv)
(\vdots)	(\vdots)

1. Differentiate the (u) column until it vanishes.
2. Integrate the (dv) column repeatedly.
3. Alternate signs (+, –, +, – …) as you pair each derivative with an integral.
4. Sum the signed products; the final term is always the last non‑zero derivative multiplied by the final integral.

Example:
[ \int x^{3}e^{x},dx ]

Derivative of (x^{3})	Integral of (e^{x})
(3x^{2})	(e^{x})
(6x)	(e^{x})
(6)	(e^{x})
(0)	(e^{x})

Now attach the alternating signs:

[ \int x^{3}e^{x},dx = x^{3}e^{x} - 3x^{2}e^{x} + 6x e^{x} - 6 e^{x} + C. ]

No repeated “boomerang” algebra is required; the table collapses everything in one sweep.

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Definite Integrals and the Limits

Integration by parts also works for definite integrals ([a,b]). The formula becomes

[ \int_{a}^{b} u,dv = \Bigl[uv\Bigr]{a}^{b} - \int{a}^{b} v,du . ]

Notice the boundary term (\bigl[uv\bigr]_{a}^{b}) appears explicitly, so you must evaluate (u) and (v) at the limits before discarding the remaining integral. This is especially useful when the antiderivative of the remaining piece is messy, but the product (uv) simplifies nicely at the endpoints.

Example:
[ \int_{0}^{\pi} x\sin x,dx. ]

Take (u=x), (dv=\sin x,dx) → (du=dx), (v=-\cos x).

[ \int_{0}^{\pi} x\sin x,dx = \bigl[-x\cos x\bigr]{0}^{\pi} + \int{0}^{\pi} \cos x,dx = \bigl[-\pi\cos\pi + 0\cdot\cos0\bigr] + \bigl[\sin x\bigr]_{0}^{\pi} = \pi + 0 = \pi . ]

The boundary term handled the “hard part,” leaving a trivial integral.

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When Integration by Parts Fails (and What to Do)

There are integrals where neither factor simplifies enough to make the method worthwhile. In such cases:

1. Try a different substitution that rewrites the integrand in a more amenable form.

2. Consider series expansion or special functions if a closed elementary antiderivative does not exist

3. Use Integration by Parts in reverse (often called "circular" or "boomerang" integration). This occurs when you encounter an integrand that repeats itself after two applications of the method, such as $\int e^x \cos x , dx$. By treating the resulting integral as another "unknown" to be solved algebraically, you can isolate the original integral on one side of the equation.

Summary and Best Practices

Mastering integration by parts is less about memorizing a single formula and more about developing an intuition for the "weight" of your terms. To use the method effectively, keep these strategic tips in mind:

• The LIATE Rule: When choosing your $u$, follow this hierarchy to ensure the derivative is simpler than the original: Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential.
• Simplify First: Always check if a simple $u$-substitution can reduce the complexity of the integrand before committing to the more labor-intensive integration by parts.
• Watch the Signs: The most common error in this method is a sign error during the subtraction of the second integral. Using the tabular method (as shown above) is an excellent way to mitigate this risk.

Conclusion

Integration by parts is a powerful tool that transforms the product of two functions into a more manageable form by leveraging the inverse relationship between differentiation and integration. Whether you are dealing with polynomial-exponential products using the tabular method or solving complex definite integrals via the boundary term, the method provides a bridge across many mathematical gaps. By recognizing when to apply it—and knowing when to pivot to substitution or series expansion—you gain complete control over the landscape of calculus.