Integration by Parts for JEE Main: Complete Guide

Integration by parts handles integrals that no substitution can crack — products of functions from different families. The formula ∫u dv = uv − ∫v du transforms the original integral into one that is (ideally) simpler. The art is choosing u correctly, and the LIATE rule makes this systematic. This guide covers every form JEE Main tests, from single application to the cyclic trap that rewards pattern recognition.

The LIATE Rule for Choosing u

The priority order for u is: Logarithmic > Inverse trig > Algebraic > Trigonometric > Exponential. Choose the function appearing earlier in this list as u. Rationale: the earlier functions become simpler when differentiated, ensuring ∫v du is easier than the original. In ∫x ln x dx: ln x is L, x is A — u = ln x, dv = x dx. In ∫x e^x dx: x is A, e^x is E — u = x, dv = e^x dx. In ∫e^x sin x dx: sin x is T, e^x is E — u = sin x (but this leads to the cyclic case, discussed below). For the integration techniques context see our integration techniques guide.

Worked Example: Single Application

∫x e^x dx: u = x (A), dv = e^x dx (E). du = dx, v = e^x. Result: xe^x − ∫e^x dx = xe^x − e^x + C = e^x(x − 1) + C. Verify by differentiating: d/dx[e^x(x−1)] = e^x(x−1) + e^x = xe^x = the integrand ✓. The verification step takes five seconds and catches coefficient errors before they cost exam marks.

The DI Tabular Method for Repeated Application

For ∫x²e^x dx, apply parts twice. The DI method is a bookkeeping shortcut: column D (differentiate u repeatedly until zero): x², 2x, 2, 0. Column I (integrate dv repeatedly): e^x, e^x, e^x, e^x. Multiply diagonally with alternating signs (+, −, +, −): x²·e^x − 2x·e^x + 2·e^x = e^x(x² − 2x + 2) + C. This diagram prevents sign errors in multi-step parts and saves 30-45 seconds per problem compared to doing two sequential part-integrations.

Cyclic Integrals: The Algebraic Trick

For ∫e^x sin x dx: applying parts twice returns the original integral. Let I = ∫e^x sin x dx. First application (u = sin x): I = e^x sin x − ∫e^x cos x dx. Second application to ∫e^x cos x dx (u = cos x): = e^x cos x + ∫e^x sin x dx = e^x cos x + I. Substituting: I = e^x sin x − e^x cos x − I. Solving: 2I = e^x(sin x − cos x), so I = ½e^x(sin x − cos x) + C. This cyclic approach avoids infinite regression and is the definitive pattern for e^x times trig.

The e^x[f(x) + f'(x)] Shortcut

∫e^x[f(x) + f'(x)] dx = e^x f(x) + C — recognised by inspection without applying parts. This arises from the product rule: d/dx[e^x f(x)] = e^x f(x) + e^x f'(x). JEE uses this in harder integration questions disguised as ∫e^x(x/(1+x)²) dx = ∫e^x[1/(1+x) − 1/(1+x)²] dx, where f(x) = 1/(1+x). Recognising f + f' is the signature of this shortcut. After practising single, repeated, cyclic, and shortcut integrations, take a free mock test on indefinite integrals.