# Integration by parts on a short product

Integration by parts comes from the product rule. If (u(x)) and (v(x)) are differentiable, then

[ \int u,dv = uv - \int v,du. ]

The usual strategy is to let (u) be the factor that gets simpler after differentiation (a polynomial, a logarithm, or an inverse trig function) and to let (dv) be a factor that is easy to integrate (an exponential, a sine or cosine, or a power with a simple antiderivative).

# Worked example

Compute (\int x e^{2x},dx).

Let (u = x), so (du = dx). Let (dv = e^{2x},dx), so (v = \frac{1}{2}e^{2x}). Then

[ \int x e^{2x},dx = \frac{x}{2}e^{2x} - \int \frac{1}{2}e^{2x},dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C = \frac{e^{2x}}{4}(2x - 1) + C. ]

Differentiate to check: the product rule on (\frac{e^{2x}}{4}(2x-1)) gives (\frac{2e^{2x}}{4}(2x-1) + \frac{e^{2x}}{4}\cdot 2 = e^{2x}(x - 1/2 + 1/2) = x e^{2x}).

A second example is (\int \ln x,dx). Take (u = \ln x) and (dv = dx). Then (du = dx/x) and (v = x), so the integral is (x\ln x - \int dx = x\ln x - x + C).

If you get stuck choosing (u) and (dv), a free step-by-step solver is at MathsSolverGPT. After you have an antiderivative, differentiating it is the most reliable check.