--- title: "Integration by Parts" description: "Reverse the product rule to evaluate integrals. Choose useful parts, repeat the calculation when needed, and solve for a returning integral." canonical: https://lightmysky.com/learn/mathematics/integration-by-parts-mt_v6jZVs2_tf source: https://lightmysky.com/learn/mathematics/integration-by-parts-mt_v6jZVs2_tf.md retrieved: 2026-11-12 --- > **Agent view.** This is the Markdown twin of the page, for tools and assistants. > When to use this site, and the call that answers each job: https://lightmysky.com/agent-instructions.md > API description (OpenAPI 3.1): https://lightmysky.com/openapi.json · Authentication: https://lightmysky.com/auth.md > Pricing: https://lightmysky.com/pricing.md · Catalog: https://lightmysky.com/llms.txt · Full catalog: https://lightmysky.com/llms-full.txt > Every machine-readable file on this domain: https://lightmysky.com/.well-known/ai-catalog.json > Ask for Markdown with `Accept: text/markdown`, a `.md` address, or `?mode=agent`. # Integration by Parts Reverse the product rule to evaluate integrals. Choose useful parts, repeat the calculation when needed, and solve for a returning integral. Subject: Mathematics · Area: Calculus & Analysis · Ages 19 to 20 Page: https://lightmysky.com/learn/mathematics/integration-by-parts-mt_v6jZVs2_tf ## Ready when they can - Choose useful parts and apply the formula to indefinite integrals or definite function data - Apply integration by parts repeatedly, including solving for an integral that reappears - Integrate logarithms and logarithmic powers by taking dv = dx, and explain the resulting cancellations ## Lesson: Trade one integral for an easier one Substitution reverses a chain-rule pattern. Integration by parts uses the product rule instead. For u and v with continuous derivatives, (uv)' = u'v+uv'. Integrating and rearranging gives ∫u dv = uv-∫v du, where du = u' dx and dv = v' dx. Choose an antiderivative v you can find, then examine the remaining integral. Keep one antiderivative during the calculation and add C at the end. **Example.** For ∫(x+3)e^x dx, take u = x+3 and dv = e^x dx. Then du = dx and v = e^x, giving (x+3)e^x-∫e^x dx = (x+2)e^x+C. Differentiating the polynomial lowers its degree. Reversing this split raises the degree in the remaining integral. For a ≠ 0, ∫e^(ax) dx = e^(ax)/a+C. LIATE orders logarithms, inverse trig, algebra, trig and exponentials as possible choices for u. It is a guide; check the resulting integral. For x > 0, the derivative of ln x is 1/x. In ∫x^n ln x dx, with n a nonnegative integer, choose u = ln x and dv = x^n dx. Then v = x^(n+1)/(n+1), so v du = x^n dx/(n+1). The remaining integral uses the power rule and contains no logarithm. For n ≥ 1, choosing the polynomial as u instead leaves a logarithm in the remaining integral. Use radians for trigonometric arguments. The antiderivatives are ∫cos x dx = sin x+C and ∫sin x dx = -cos x+C. For a polynomial p, parts gives ∫p(x)cos x dx = p(x)sin x-∫p'(x)sin x dx. With sine instead, v = -cos x, so subtracting ∫v du adds ∫p'(x)cos x dx. Check the signs by differentiating the complete answer using the product rule. **Example.** For x > 0, integrate ln x by taking u = ln x and dv = dx. Then du = dx/x and v = x. Thus ∫ln x dx = x ln x-∫1 dx = x ln x-x+C. For definite bounds a and b, use F(b)-F(a), where F(x) = x ln x-x. For example, from 1 to e² the endpoint values are e² and -1, giving e²+1. The arbitrary constant cancels. Let R_n stand for ∫x^n e^(ax) dx, with n a nonnegative integer and a ≠ 0. For n ≥ 1, take u = x^n and dv = e^(ax) dx. Then R_n = x^n e^(ax)/a-(n/a)R_(n-1), up to a constant. Repeat on the remaining integral. Each application lowers the polynomial degree by one. After n applications, integrate the remaining constant times e^(ax) directly. Keep each factor 1/a and each outside subtraction. Let I = ∫e^(ax)cos x dx and J = ∫e^(ax)sin x dx, with a ≠ 0. Taking the trig factor as u gives I = e^(ax)cos x/a+J/a and J = e^(ax)sin x/a-I/a, up to constants. Substitute the expression for J into I. Collecting terms gives (a²+1)I = e^(ax)(a cos x+sin x). Divide by a²+1 and add C. A returning integral can be solved with algebra. You can also start with the sine product. Let S = ∫e^(ax)sin x dx and T = ∫e^(ax)cos x dx, with a ≠ 0. Since (sin x)' = cos x, parts gives S = e^(ax)sin x/a-T/a. Since (cos x)' = -sin x, the next application gives T = e^(ax)cos x/a+S/a. Substitute the whole expression for T, including its plus sign, into the subtracted term. For u and v with continuous derivatives on [a,b], definite parts is ∫ from a to b of u dv = u(b)v(b)-u(a)v(a)-∫ from a to b of v du. Supplied endpoint values and a remaining integral can be enough, even without a formula for the function. For u = mx+c and dv = f'(x) dx, use du = m dx and v = f(x). Evaluate the upper product minus the lower product, then subtract m∫f(x) dx over the same bounds. For x > 0, powers of ln x can need repeated parts. With a positive integer m, choose u = (ln x)^m and dv = dx. Then du = m(ln x)^(m-1) dx/x and v = x. Cancelling x gives ∫(ln x)^m dx = x(ln x)^m-m∫(ln x)^(m-1) dx, up to a constant. Repeat until the remaining integral is ∫ln x dx or ∫1 dx. Distribute every outside minus sign. Any antiderivative of dv may be used as v. For ∫ln x dx on x > 0, choosing v = x+k still gives dv = dx, where k is constant. Now v du = (1+k/x) dx. Parts gives (x+k)ln x-x-k ln x+C. The added logarithmic terms cancel, leaving x ln x-x+C. Choosing a constant in v does not change the final family of antiderivatives. *(drawing: For x³ times e^x, three parts applications leave a constant times e^x. Integrate that last exponential directly.)* **Recap.** Choose an integrable dv and compute uv-∫v du. Repeat or solve for a returning integral when needed. Evaluate definite bounds in order and check antiderivatives by differentiating. *Integration-by-parts teaching, practice questions and flashcards written for this learning dataset.* ## Practice 14 questions on this page, each with its working shown. ## Needs first - [Integration as the Reverse of Differentiation](https://lightmysky.com/learn/mathematics/integration-as-the-reverse-of-differentiation-mt_edtk3ArxRk) ## Optional connections - [The Product and Quotient Rules](https://lightmysky.com/learn/mathematics/the-product-and-quotient-rules-mt_NJ7INJ0jfv) ## Opens up - [Weak Solutions and Test Functions](https://lightmysky.com/learn/mathematics/weak-solutions-and-test-functions-mt_AKlC3Ol70p)