Integration by Parts

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.

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.

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.

Here A = e^(ax)cos x/a and B = e^(ax)sin x/a.

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.