Integration by Parts Essentials: From Definition to First Tricks

To solve ∫ x eˣ dx using integration by parts, we use the formula ∫ u dv = uv - ∫ v du. Let u = x and dv = eˣ dx. Then du = dx and v = eˣ. Substituting into the formula gives us ∫ x eˣ dx = x eˣ - ∫ eˣ dx. The remaining integral ∫ eˣ dx is simply eˣ + C. Therefore, the final answer is x eˣ - eˣ + C.

We apply integration by parts with u = x and dv = cos x dx. This gives du = dx and v = sin x. Using the formula ∫ u dv = uv - ∫ v du, we get ∫ x cos x dx = x sin x - ∫ sin x dx. The integral of sin x is -cos x + C. So we have x sin x - (-cos x) + C, which simplifies to x sin x + cos x + C.

First find the antiderivative using integration by parts. Let u = x and dv = eˣ dx, so du = dx and v = eˣ. Then ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C = eˣ (x - 1) + C. Now evaluate from 0 to 1: [e^1 (1 - 1)] - [e⁰(0 - 1)] = [e × 0] - [1 × (-1)] = 0 - (-1) = 1.

For ∫ ln x dx, we need to choose u = ln x and dv = dx. This is because the derivative of ln x is 1/x, which simplifies the problem, while the integral of dx is simply x. If we chose u = x and dv = ln x dx, we would need to find the antiderivative of ln x, which is what we're trying to compute in the first place. The standard approach is to let u = ln x and dv = dx.

First application: Let u = x² and dv = eˣ dx, so du = 2x dx and v = eˣ. Then ∫ x² eˣ dx = x² eˣ - ∫ 2x eˣ dx. Now we need to evaluate ∫ x eˣ dx (the factor 2 can be pulled out). For this, apply integration by parts again: let u = x and dv = eˣ dx, giving du = dx and v = eˣ. So ∫ x eˣ dx = x eˣ - ∫ eˣ dx = x eˣ - eˣ + C. Putting it all together: ∫ x² eˣ dx = x² eˣ - 2(x eˣ - eˣ) + C = x² eˣ - 2x eˣ + 2eˣ + C = eˣ (x² - 2x + 2) + C.

If we choose u = sin x and dv = x dx, then du = cos x dx and v = x²/2. Applying integration by parts gives ∫ x sin x dx = (sin x)(x²/2) - ∫ (x²/2) cos x dx. The new integral ∫ (x²/2) cos x dx is more complicated than the original because it involves x² instead of x. This is why we typically choose u = x and dv = sin x dx for this integral, as this choice reduces the power of x and simplifies the problem.

First find the antiderivative. Let u = x and dv = sin x dx, so du = dx and v = -cos x. Then ∫ x sin x dx = x(-cos x) - ∫ (-cos x) dx = -x cos x + ∫ cos x dx = -x cos x + sin x + C. Now evaluate from 0 to π: [-(π)cos(π) + sin(π)] - [-(0)cos(0) + sin(0)] = [-π × (-1) + 0] - [0 + 0] = π.

Let u = x and dv = e²ˣ dx. Then du = dx and v = (½)e²ˣ (since the integral of e²ˣ is (½)e²ˣ). Applying integration by parts: ∫ x e²ˣ dx = x × (½)e²ˣ - ∫ (½)e²ˣ dx = (x/2)e²ˣ - (½) × (½)e²ˣ + C = (x/2)e²ˣ - (1/4)e²ˣ + C.

The integration by parts formula is derived from the product rule for differentiation. The product rule states that d(uv) = u dv + v du. Integrating both sides gives uv = ∫ u dv + ∫ v du. Rearranging this equation to solve for ∫ u dv yields the integration by parts formula: ∫ u dv = uv - ∫ v du.

Let u = ln x and dv = x³ dx. Then du = (1/x) dx and v = x⁴/4. Using integration by parts: ∫ x³ ln x dx = (ln x)(x⁴/4) - ∫ (x⁴/4)(1/x) dx = (x⁴/4)ln x - ∫ (x³/4) dx = (x⁴/4)ln x - (1/4)(x⁴/4) + C = (x⁴/4)ln x - x⁴/16 + C.

The LIATE rule is a helpful guideline for choosing u and dv. Functions near the beginning of the list (Logarithmic, Inverse trig) are generally good choices for u because their derivatives simplify, while functions near the end (Trig, Exponential) are good choices for dv because their antiderivatives are straightforward. This helps ensure that the new integral ∫ v du is simpler than the original ∫ u dv.

First application: Let u = sin x, dv = eˣ dx, so du = cos x dx, v = eˣ. Then ∫ eˣ sin x dx = eˣ sin x - ∫ eˣ cos x dx. Second application on ∫ eˣ cos x dx: Let u = cos x, dv = eˣ dx, so du = -sin x dx, v = eˣ. This gives ∫ eˣ cos x dx = eˣ cos x + ∫ eˣ sin x dx. Substituting back: ∫ eˣ sin x dx = eˣ sin x - [eˣ cos x + ∫ eˣ sin x dx] = eˣ sin x - eˣ cos x - ∫ eˣ sin x dx. Now add ∫ eˣ sin x dx to both sides: 2∫ eˣ sin x dx = eˣ(sin x - cos x). Therefore, ∫ eˣ sin x dx = (eˣ)(sin x - cos x)/2 + C.

For definite integrals, the most reliable method is to first find the antiderivative using integration by parts (including the constant of integration + C), and then apply the Fundamental Theorem of Calculus by evaluating the antiderivative at the upper limit and subtracting its value at the lower limit. This approach is clearer and reduces errors compared to trying to apply limits during the integration by parts process.

Let u = ln x and dv = x dx, so du = (1/x) dx and v = x²/2. Then ∫ x ln x dx = (x²/2)ln x - ∫ (x²/2)(1/x) dx = (x²/2)ln x - ∫ (x/2) dx = (x²/2)ln x - x²/4 + C. Now evaluate from 1 to e: [(e²/2)ln e - e²/4] - [(1²/2)ln 1 - 1²/4] = [(e²/2)(1) - e²/4] - [(½)(0) - 1/4] = (e²/2 - e²/4) - (-1/4) = (e²/4) + (1/4) = (e² + 1)/4.

Some integrals require integration by parts multiple times, especially when the integrand involves a polynomial of degree 2 or higher multiplied by an exponential or trigonometric function. Each application typically reduces the degree of the polynomial part. You continue applying the method until the remaining integral is one you can evaluate directly, such as a simple exponential or trigonometric integral. This is why repeated integration by parts is a powerful technique for integrals like ∫ x² eˣ dx or ∫ eˣ sin x dx.