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Pure · Integration

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Integration by parts

Reversing the product rule to integrate a product — ∫u(dv/dx)dx = uv − ∫v(du/dx)dx — choosing u to differentiate to something simpler, the ∫ln x trick, and applying it twice.

Pure · Integration

Integration by parts

Reversing the product rule to integrate a product — ∫u(dv/dx)dx = uv − ∫v(du/dx)dx — choosing u to differentiate to something simpler, the ∫ln x trick, and applying it twice.

Why it works

Some products can't be integrated by recognition — ∫xex dx\int x e^x\,dx has no inside function whose derivative is sitting there. Integration by parts handles these by reversing the product rule.

Start from the product rule ddx(uv)=udvdx+vdudx\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx} and integrate both sides: uv=∫udvdx dx+∫vdudx dxuv = \int u\dfrac{dv}{dx}\,dx + \int v\dfrac{du}{dx}\,dx. Rearranged, that is the formula: ∫u dvdx dx=uv−∫v dudx dx.\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx. You split the integrand into a part to differentiate (uu) and a part to integrate (dvdx\dfrac{dv}{dx}). The trade is only worth it if the new integral ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx is easier than the one you started with.

Choosing uu. Pick uu to be the factor that gets simpler when differentiated, so the leftover integral improves. A reliable order of preference for uu is L–A–T–E: Logs, then Algebra (powers of xx), then Trig, then Exponentials. So in ∫xex dx\int x e^x\,dx take u=xu = x (algebra beats exponential): differentiating xx gives 11, killing the power. ∫xex dx:u=x, dvdx=ex  ⇒  dudx=1, v=ex.\int x e^x\,dx: \quad u = x,\ \frac{dv}{dx} = e^x \;\Rightarrow\; \frac{du}{dx} = 1,\ v = e^x. =xex−∫ex⋅1 dx=xex−ex+c.= x e^x - \int e^x\cdot 1\,dx = x e^x - e^x + c.

The ∫ln⁡x dx\boldsymbol{\int \ln x\,dx} trick. There's no obvious product, but write ln⁡x=ln⁡x×1\ln x = \ln x \times 1 and take u=ln⁡xu = \ln x, dvdx=1\dfrac{dv}{dx} = 1. Then dudx=1x\dfrac{du}{dx} = \tfrac1x, v=xv = x, and ∫ln⁡x dx=xln⁡x−∫x⋅1x dx=xln⁡x−x+c.\int \ln x\,dx = x\ln x - \int x\cdot\tfrac1x\,dx = x\ln x - x + c.

Applying it twice. For ∫x2ex dx\int x^2 e^x\,dx one pass leaves ∫2xex dx\int 2x e^x\,dx, which still needs parts — so do it again. Each application drops the power of xx by one until it disappears.