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Unlocking Integrals: The Art of Integration by Parts

An intuitive look at integration by parts as the reverse of the product rule, applied to the function x e^(2x).

The Product Ruled/dx(u v) = u v' + v u'
ddx(uv)=udvdx+vdudx\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}
Integration by parts is simply the product rule of differentiation run in reverse. When we see a product of two functions, we look for a way to untangle them.
Step-by-step solver
1

(Solution) Identify parts

Set u as the algebraic part to simplify it, and dv as the exponential part to integrate.

u=x,dv=e2x dxu = x, \quad dv = e^{2x}\,dx
2

(Solution) Calculate differentials/integrals

Find du by differentiating u and v by integrating dv.

du=dx,v=12e2xdu = dx, \quad v = \tfrac{1}{2}e^{2x}
3

(Solution) Substitute into formula

Use the integration by parts formula: ∫ u dv = uv - ∫ v du.

∫xe2x dx=x⋅12e2x−∫12e2x dx\int x e^{2x}\,dx = x \cdot \tfrac{1}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx
4

(Solution) Final integration

Perform the remaining integration and add the constant of integration C.

12xe2x−12(12e2x)+C=12xe2x−14e2x+C\tfrac{1}{2}x e^{2x} - \tfrac{1}{2}(\tfrac{1}{2}e^{2x}) + C = \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + C

Original question

Find the integral of x e^(2x) with respect to x, using integration by parts. Show full working.

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