Integration by Parts and Partial Fractions
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Direct answer
Integration by parts trades one integral for another: ∫u·v dx = u∫v dx − ∫(u' ∫v dx)dx, and the technique lives or dies on choosing u wisely — by the ILATE order (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential), the priority list Indian classrooms drill. Partial fractions is the sibling tool for rational functions: a proper fraction with denominator (x − 1)(x + 2) splits as A/(x − 1) + B/(x + 2), the constants read off by strategic substitution, and each piece integrates to a logarithm. Between them, these two methods clear the majority of integration marks in JEE — by parts for mixed products, partial fractions for quotients.
What you must remember
- ILATE: the function higher on the list (Inverse, Log, Algebraic, Trig, Exponential) is chosen as u; ∫ln x dx = x ln x − x + C is the archetype of the rule paying off.
- Classic results: ∫sin^(-1) x dx = x sin^(-1) x + √(1 − x^2) + C; ∫e^x sin x dx = e^x(sin x − cos x)/2 + C, obtained by parts twice and solved algebraically.
- The e^x twin rule: ∫e^x(f(x) + f'(x))dx = e^x f(x) + C — so ∫e^x(1/x − 1/x^2)dx = e^x/x + C; spot the f, f' pattern before grinding.
- Partial fraction shapes: distinct linear factors give A/(x − a) + B/(x − b); a repeated factor (x − a)^2 needs A/(x − a) + B/(x − a)^2; an irreducible quadratic takes (Ax + B) over it.
- Improper fractions: divide first (polynomial long division) whenever the numerator's degree reaches the denominator's.
- Cover-up shortcut: for a distinct linear factor (x − a), cover it and evaluate the rest at x = a — the Heaviside cover-up that wins seconds in the hall.
- Log discipline: every linear partial-fraction term integrates to ln|...|; keep the modulus to the end.
Resolving a rational function
Compute ∫(2x + 3)/((x − 1)(x + 2)) dx. Set (2x + 3)/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2), so 2x + 3 = A(x + 2) + B(x − 1). Substitute x = 1: 5 = 3A, so A = 5/3. Substitute x = −2: −1 = −3B, so B = 1/3. The cover-up reading gives the same values instantly: cover (x − 1), evaluate (2x + 3)/(x + 2) at x = 1 to get 5/3. The integral becomes (5/3)ln|x − 1| + (1/3)ln|x + 2| + C. For contrast, one by-parts anchor: ∫x e^x dx with u = x, v' = e^x gives x e^x − ∫e^x dx = e^x(x − 1) + C — the choice of u = x (algebraic before exponential in ILATE) is what made the second integral easier than the first, and that improvement is the entire criterion by which u is chosen.
Where marks leak
JEE Main asks for a single by-parts application or a two-factor partial fraction — mechanical, provided the choices are right. JEE Advanced builds cyclic by-parts (e^x sin x, where the original integral reappears and is solved for) and partial fractions with a repeated or quadratic factor needing (Ax + B). The recurring losses: choosing the exponential as u, which makes the new integral worse instead of better; splitting a repeated factor (x − a)^2 as only A/(x − a) + B/(x − a) — the second term must carry the power; applying partial fractions to an improper fraction without dividing first; and dropping modulus signs from the logarithms. The e^x twin rule is also under-used: examiners plant ∫e^x(tan x + sec^2 x)dx precisely for candidates who will grind two by-parts applications instead of writing e^x tan x + C.
Frequently asked questions
What order does ILATE prescribe?
Inverse trigonometric, logarithmic, algebraic, trigonometric, exponential — the earlier function is chosen as u in ∫u·v dx.
What is ∫e^x(f(x) + f'(x))dx?
e^x f(x) + C, the twin rule that turns a long by-parts computation into recognition.
How does (2x + 3)/((x − 1)(x + 2)) split?
Into 5/3 over (x − 1) plus 1/3 over (x + 2), by substituting x = 1 and x = −2.
What does a repeated factor (x − a)^2 require?
Both A/(x − a) and B/(x − a)^2 terms — one constant per power, down to the first.
What must precede partial fractions on an improper rational function?
Polynomial long division, producing a polynomial plus a proper fraction that can then be split.