Integration by Parts with Polynomials, Logs, and Definite Integrals

Let u = 2x + 3 and dv = eˣ dx. Then du = 2 dx and v = eˣ. Applying integration by parts: ∫ (2x + 3) eˣ dx = (2x + 3)eˣ - ∫ 2eˣ dx = (2x + 3)eˣ - 2eˣ + C = eˣ (2x + 3 - 2) + C = eˣ (2x + 1) + C. So the answer simplifies to (2x + 1)eˣ + C.

Let u = x and dv = sin(3x) dx. Then du = dx and v = -(⅓) cos(3x). Applying integration by parts: ∫ x sin(3x) dx = x × [-(⅓) cos(3x)] - ∫ [-(⅓) cos(3x)] dx = -(x/3) cos(3x) + (⅓) ∫ cos(3x) dx. The integral of cos(3x) is (⅓) sin(3x). So we get -(x/3) cos(3x) + (⅓) × (⅓) sin(3x) + C = -(x/3) cos(3x) + (1/9) sin(3x) + C.

The main goal when selecting u and dv is to ensure that the resulting integral ∫ v du is simpler than the original integral ∫ u dv. Typically, we want u to be a function that becomes simpler when differentiated (like a polynomial), and dv to be a function that is easy to integrate (like eˣ or sin x). This strategic choice is what makes integration by parts effective.

First find the antiderivative: ∫ (x + 1) eˣ dx. Let u = x + 1, dv = eˣ dx → du = dx, v = eˣ. Then ∫ (x + 1) eˣ dx = (x + 1)eˣ - ∫ eˣ dx = (x + 1)eˣ - eˣ + C = x eˣ + C. Now evaluate from 0 to 1: [1 × e^1] - [0 × e^0] = e - 0 = e.

First, simplify ln(x²) = 2 ln x. So ∫ x³ ln(x²) dx = ∫ x³ (2 ln x) dx = 2 ∫ x³ ln x dx. Now apply integration by parts to ∫ x³ ln x dx: Let u = ln x, dv = x³ dx → du = (1/x) dx, v = x⁴/4. Then ∫ x³ ln x dx = (x⁴/4) ln x - ∫ (x⁴/4)(1/x) dx = (x⁴/4) ln x - ∫ (x³/4) dx = (x⁴/4) ln x - x⁴/16 + C. Multiply by 2: 2[(x⁴/4) ln x - x⁴/16] = (x⁴/2) ln x - x⁴/8 + C. Since ln(x²) = 2 ln x, we can write (x⁴/2) ln x = (x⁴/4) ln(x²). So the answer is (x⁴/4) ln(x²) - x⁴/8 + C.

First application: u = x², dv = cos x dx → du = 2x dx, v = sin x. Then ∫ x² cos x dx = x² sin x - ∫ 2x sin x dx. Second application on ∫ x sin x dx: u = x, dv = sin x dx → du = dx, v = -cos x. This gives ∫ x sin x dx = -x cos x + ∫ cos x dx = -x cos x + sin x + C. So ∫ x² cos x dx = x² sin x - 2(-x cos x + sin x) + C = x² sin x + 2x cos x - 2 sin x + C.

The constant of integration (+ C) is necessary because when we differentiate any constant, we get zero. This means that if F(x) is an antiderivative of f(x), then F(x) + C is also an antiderivative for any constant C. The indefinite integral represents a family of functions, all differing by a constant, that have the same derivative. We include + C to represent this entire family of solutions.

Let u = ln x and dv = dx, so du = (1/x) dx and v = x. Then ∫ ln x dx = x ln x - ∫ x × (1/x) dx = x ln x - ∫ dx = x ln x - x + C. Now evaluate from 1 to e: [e ln e - e] - [1 ln 1 - 1] = [e × 1 - e] - [0 - 1] = (e - e) - (-1) = 0 + 1 = 1.

First application: Let u = sin x, dv = e²ˣ dx → du = cos x dx, v = (½)e²ˣ. Then ∫ e²ˣ sin x dx = (½)e²ˣ sin x - ∫ (½)e²ˣ cos x dx. Second application on ∫ e²ˣ cos x dx: Let u = cos x, dv = e²ˣ dx → du = -sin x dx, v = (½)e²ˣ. This gives ∫ e²ˣ cos x dx = (½)e²ˣ cos x + ∫ (½)e²ˣ sin x dx. Substituting back: ∫ e²ˣ sin x dx = (½)e²ˣ sin x - (½)[(½)e²ˣ cos x + (½)∫ e²ˣ sin x dx] = (½)e²ˣ sin x - (1/4)e²ˣ cos x - (1/4)∫ e²ˣ sin x dx. Bring the integral to the left side: (1 + 1/4)∫ e²ˣ sin x dx = (½)e²ˣ sin x - (1/4)e²ˣ cos x. So (5/4)∫ e²ˣ sin x dx = e²ˣ[(½) sin x - (1/4) cos x]. Multiply both sides by 4/5: ∫ e²ˣ sin x dx = e²ˣ[(2/5) sin x - (1/5) cos x] + C = (e²ˣ)(2 sin x - cos x)/5 + C.

Integration by parts is used when the integrand is a product of two functions that are not related by differentiation. U-substitution works when one part of the product is the derivative of the inner function (chain rule scenario). Integration by parts works for products like x eˣ or x sin x where one function is not the derivative of the other.

First find the antiderivative using integration by parts twice. First: u = x², dv = sin x dx → du = 2x dx, v = -cos x. Then ∫ x² sin x dx = -x² cos x + ∫ 2x cos x dx. Second: For ∫ x cos x dx, u = x, dv = cos x dx → du = dx, v = sin x. This gives ∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x. So ∫ x² sin x dx = -x² cos x + 2(x sin x + cos x) + C = -x² cos x + 2x sin x + 2 cos x + C. Evaluate from 0 to π/2: [-(π/2)² cos(π/2) + 2(π/2) sin(π/2) + 2 cos(π/2)] - [-(0)² cos(0) + 2(0) sin(0) + 2 cos(0)] = [-(π²/4) × 0 + π × 1 + 2 × 0] - [0 + 0 + 2 × 1] = π - 2.

The integral ∫ eˣ sin x dx requires a special technique. After applying integration by parts twice, you end up with the original integral appearing again on the right side. This creates an equation where you can solve for the original integral algebraically. This circular approach is necessary because the exponential and trigonometric functions cycle through differentiation and integration without simplifying.

Applying integration by parts repeatedly: ∫ x⁴ eˣ dx = x⁴ eˣ - 4 ∫ x³ eˣ dx. Similarly, ∫ x³ eˣ dx = x³ eˣ - 3 ∫ x² eˣ dx. And ∫ x² eˣ dx = x² eˣ - 2 ∫ x eˣ dx. And ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ. Working backwards: ∫ x² eˣ dx = x² eˣ - 2(x eˣ - eˣ) = x² eˣ - 2x eˣ + 2eˣ. ∫ x³ eˣ dx = x³ eˣ - 3(x² eˣ - 2x eˣ + 2eˣ) = x³ eˣ - 3x² eˣ + 6x eˣ - 6eˣ. Finally, ∫ x⁴ eˣ dx = x⁴ eˣ - 4(x³ eˣ - 3x² eˣ + 6x eˣ - 6eˣ) + C = x⁴ eˣ - 4x³ eˣ + 12x² eˣ - 24x eˣ + 24eˣ + C = eˣ (x⁴ - 4x³ + 12x² - 24x + 24) + C.

While choosing u and dv strategically is important, you can sometimes recover from a less-than-optimal choice by applying integration by parts again or adjusting your approach. The other statements are true: integration by parts works for both definite and indefinite integrals, can be applied multiple times (as seen with repeated integration by parts), and is derived from the product rule.

The product rule states d/dx(uv) = u'v + uv'. Integrating both sides gives uv = ∫ u'v dx + ∫ uv' dx. Rearranging yields ∫ uv' dx = uv - ∫ u'v dx. Letting dv = v' dx and du = u' dx gives the integration by parts formula ∫ u dv = uv - ∫ v du.