Integrals
GradesPre-Calculus & CalculusIntegralsIntegration by Parts

Integration by Parts

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Integration by Parts

Integrate products of functions using the reverse of the product rule.

Integration by parts is derived from the product rule. It converts the integral of a product into a simpler integral. The formula is ∫u dv = uv − ∫v du. Choose u and dv strategically.

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
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LIATE rule for choosing u: Logarithms, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Choose the first type that appears as u.

Integration by Parts

1Identify u and dv using the LIATE rule.
2Find du by differentiating u.
3Find v by integrating dv.
4Substitute into ∫u dv = uv − ∫v du.
5Evaluate the remaining integral (may need to apply IBP again).
✏️Evaluate ∫ x·eˣ dx
LIATE: u = x (algebraic), dv = eˣ dx. Then du = dx, v = eˣ. Apply formula: ∫x eˣ dx = x·eˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C.
∫xex dx=xex−ex+C=ex(x−1)+C\int x e^x\,dx = xe^x - e^x + C = e^x(x-1) + C
✏️Evaluate ∫ x·sin(x) dx
u = x, dv = sin(x) dx → du = dx, v = −cos(x). ∫x sin x dx = −x cos x − ∫(−cos x) dx = −x cos x + sin x + C.
∫xsin⁡x dx=−xcos⁡x+sin⁡x+C\int x\sin x\,dx = -x\cos x + \sin x + C
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Remember This!

For ∫ ln(x) dx, use u = ln(x), dv = dx (so v = x): ∫ ln x dx = x ln x − x + C. Integration by parts works even when there is no obvious second factor.

✏️ Try It!

Using integration by parts, which is the correct result for ∫ x²eˣ dx (applied twice)?

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