Integration by Parts

Split the integrand into two factors u and v′. Differentiate u (giving u′) and integrate v′ (giving v). Then insert into the formula ∫u·v′ dx = u·v − ∫u′·v dx and solve the new, hopefully simpler integral. Example ∫x·eˣ dx: choose u = x, v′ = eˣ, so u′ = 1, v = eˣ. Substituting: x·eˣ − ∫1·eˣ dx = x·eˣ − eˣ + C = (x − 1)eˣ + C. Check by differentiating: eˣ + (x − 1)eˣ = x·eˣ ✓. The method is called "by parts" because only part of the product is integrated; the rest is shifted into a new integral, which must be easier to solve, otherwise the split was badly chosen.

Guiding idea: u should become simpler when differentiated, v′ must be easy to integrate. Polynomials are ideal u candidates, since their degree drops with every differentiation: x becomes 1, and the remaining integral becomes trivial. Exponential, sine and cosine are ideal v′ candidates, because integrating them does not make them more complicated. So for ∫x·sin x dx choose u = x, v′ = sin x. Logarithms are the exception: ln x is hard to integrate directly but easy to differentiate, so ln x always becomes u, if necessary with the trick ∫ln x dx = ∫1·ln x dx. The priority list LIATE (Logarithmic, Inverse, Algebraic, Trigonometric, Exponential) serves as a mnemonic: whatever stands further left becomes u.

With the 1 trick: write ln x as the product 1·ln x. Choose u = ln x (easy to differentiate) and v′ = 1 (easy to integrate), so u′ = 1/x and v = x. The formula gives ∫ln x dx = x·ln x − ∫x·(1/x) dx = x·ln x − ∫1 dx = x·ln x − x + C. Check: (x·ln x − x)′ = ln x + x·(1/x) − 1 = ln x ✓. The trick works whenever a function is hard to integrate but easy to differentiate; the antiderivatives of arctan x or arcsin x are found the same way. In exams ∫ln x dx is a classic, often part of an area task with the ln curve, and without the 1 trick you cannot even start.

Then apply the method again. For ∫x²·eˣ dx the first round (u = x²) lowers the degree to ∫2x·eˣ dx, the second (u = 2x) to ∫2·eˣ dx, which is elementary. Result: (x² − 2x + 2)eˣ + C. It is important to keep the roles consistent: always differentiate the polynomial, otherwise you undo the first step. A special case is the "phoenix": for ∫eˣ·sin x dx the original integral I itself reappears after two rounds, with factor −1. You solve the resulting equation I = eˣ(sin x − cos x) − I algebraically: 2I = eˣ(sin x − cos x), so I = ½eˣ(sin x − cos x) + C.

It is the integrated product rule. The product rule says (u·v)′ = u′·v + u·v′. Integrating both sides over x, the left side becomes u·v (integration undoes differentiation) and the right side the sum of the two integrals: u·v = ∫u′·v dx + ∫u·v′ dx. Solving for the second integral produces the familiar formula ∫u·v′ dx = u·v − ∫u′·v dx. This understanding is more than theory: it explains why the minus sign is there (it comes from rearranging), why an "integrated-out" boundary term u·v appears, and why the method is the counterpart of substitution, which in turn reverses the chain rule. For definite integrals the boundary term becomes [u·v]ₐᵇ evaluated at the limits.