Quantitative Aptitude · Differential and Integral Calculus

Integration by Parts and Partial Fractions for CA Foundation

Updated 1 October 2026 · Fact-checked

Integration by parts integrates a product of two functions using ∫u·v dx = u∫v dx − ∫(u′∫v dx) dx, choosing u by ILATE. Partial fractions splits a rational function into simple fractions that integrate to logarithms. Pick the method from the shape of the integrand, then check by differentiating an option.

Understand Integration by Parts and Partial Fractions

Some integrals are not in the standard table. Two common types in CA Foundation are a product of two different kinds of functions, such as x·eˣ, and a rational function, such as 1/((x−1)(x−2)).

Integration by parts comes from the product rule of differentiation. You split the integrand into two parts. One part, u, you differentiate. The other part, v, you integrate. The integral then becomes simpler, because the new integral has u′ in it, which is usually easier than u.

The skill is choosing u. Use ILATE, which gives the order of preference for u: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. The function that comes first in the list becomes u. For x·eˣ, x is Algebraic and eˣ is Exponential, so u = x.

Partial fractions is for a rational function, which is a polynomial divided by a polynomial. You break it into simple pieces such as A/(x−a) + B/(x−b). Each piece integrates to A·ln|x−a| and so on. The method works directly only when the fraction is proper, meaning the degree of the numerator is less than the degree of the denominator. If it is not, divide first.

In an MCQ paper you do not need to show working. You need to recognise the type fast, get the answer, and avoid sign slips.

Key formulas to remember

Integration by parts: ∫u·v dx = u·∫v dx − ∫( u′ · ∫v dx ) dx; u is the function chosen by ILATE. v is the other one. Do not forget the minus sign.
ILATE order: Inverse trig → Logarithmic → Algebraic → Trigonometric → Exponential; The function that appears earlier in this list is taken as u.
Integral of log x: ∫log x dx = x·log x − x + C; Here log means natural log (base e). Take u = log x and v = 1.
Integral of x·eˣ: ∫x·eˣ dx = eˣ(x − 1) + C; A standard result worth remembering for MCQs.
Special eˣ form: ∫eˣ[f(x) + f′(x)] dx = eˣ·f(x) + C; Use it when the bracket is a function plus its own derivative.
Distinct linear factors: P(x)/((x−a)(x−b)) = A/(x−a) + B/(x−b); Valid when the fraction is proper and a ≠ b. Find A and B by putting x = a and x = b.
Repeated linear factor: P(x)/((x−a)²(x−b)) = A/(x−a) + B/(x−a)² + C/(x−b); A repeated factor needs one term for each power, up to the highest power.
Basic log integral: ∫1/(x−a) dx = log|x−a| + C; Also ∫1/(x−a)² dx = −1/(x−a) + C.
Difference of squares form: ∫1/(x² − a²) dx = (1/2a)·log|(x−a)/(x+a)| + C; Valid for a ≠ 0, on any interval where x ≠ ±a (the integrand is undefined at x = ±a). This is a ready result from partial fractions.

How to solve Integration by Parts and Partial Fractions questions

First decide whether the integrand is a product of different types of function or a rational function. Then follow the matching steps.

1Check if the integral is in the standard table or solvable by simple substitution. If yes, use that and stop.
2If it is a product like x·eˣ, x·log x or x·sin x, use by parts. Choose u by ILATE and let the rest be v.
3Write u, u′, v and ∫v dx separately on rough paper. Then apply ∫u·v dx = u∫v dx − ∫(u′∫v dx) dx.
4If the new integral is still a product, apply by parts again. Keep the signs carefully.
5If the integrand is a rational function, first check it is proper. If the numerator degree is equal or higher, divide to get a polynomial plus a proper fraction.
6Factorise the denominator and write the partial fraction form with unknowns A, B, C. Multiply through by the denominator and put in convenient values of x to find them.
7Integrate each simple fraction using ∫1/(x−a) dx = log|x−a|. Combine the logs and add C.
8Match your result with the options. If unsure, differentiate an option and see if you get the original integrand.

Quickest way: Spot the type, use cover-up, verify by differentiation

When to use it: Use this in the MCQ paper when you have about 1.5 to 2 minutes per question and the options are close to each other.

Look at the integrand for 3 seconds. A product of x and eˣ, log x or a trig function means by parts. A fraction with a factorised denominator means partial fractions.
For by parts, apply ILATE in your head. Remember x·eˣ gives eˣ(x−1) and x·log x gives (x²/2)log x − x²/4 so you can recognise them fast.
For distinct linear factors, use the cover-up method. To find A for (x−a), cross out (x−a) in the denominator and put x = a in what remains.
Eliminate options using the sign and the form. A by-parts answer always has two terms. A partial-fraction answer with distinct factors is a combination of logs.
If two options remain, differentiate each one quickly. Only one will give the original integrand.
If the integral needs three rounds of by parts or a long division and you are not sure, skip it. A wrong answer costs 0.25 marks.

Common mistakes in Integration by Parts and Partial Fractions

Choosing u wrongly, for example taking eˣ as u in x·eˣ.
Students pick the function that looks easier to integrate without using ILATE.
Fix: Always apply ILATE. The function earlier in the list is u. Check that differentiating u makes the new integral simpler.
Dropping the minus sign in front of ∫(u′∫v dx) dx.
Students rush and copy the product rule from memory as a plus.
Fix: Say it as 'u times integral of v, minus integral of derivative of u times integral of v'. Write the minus first, then fill in the rest.
Applying partial fractions to an improper fraction directly.
Students see a factorised denominator and start splitting at once.
Fix: Compare the degrees first. If the numerator degree is equal or more, divide to get quotient plus a proper fraction, then split only the proper part.
Getting A and B right but integrating 1/(x−a) as 1/(x−a)² or as x−a.
Students mix up the standard integral with the power rule.
Fix: Remember that 1/(x−a) always gives log|x−a|. The power rule gives log only for the power −1.
Missing the term for a repeated factor, such as using only A/(x−1) for (x−1)².
Students treat the repeated factor as a single factor.
Fix: Write one term for each power: A/(x−1) + B/(x−1)². The number of unknowns equals the degree of the denominator.
Swapping the log ratio, writing log|(x−1)/(x−2)| instead of log|(x−2)/(x−1)|.
A and B have opposite signs, and students lose track when combining the two logs.
Fix: Keep the logs separate as A·log|x−a| + B·log|x−b| until the end. Then combine using log m − log n = log(m/n).

Worked examples

Example 1

∫x·eˣ dx equals: (a) eˣ(x − 1) + C (b) eˣ(x + 1) + C (c) x·eˣ + C (d) (x²/2)·eˣ + C

Show the solution

By ILATE, x is Algebraic and eˣ is Exponential. So u = x and v = eˣ.
u′ = 1 and ∫v dx = eˣ.
Apply the formula: ∫x·eˣ dx = x·eˣ − ∫1·eˣ dx.
The remaining integral is eˣ. So the result is x·eˣ − eˣ + C = eˣ(x − 1) + C.
Check by differentiating: d/dx [eˣ(x − 1)] = eˣ(x − 1) + eˣ = x·eˣ. This matches.

Answer: (a) eˣ(x − 1) + C

Example 2

∫1/((x − 1)(x − 2)) dx equals: (a) log|(x − 1)/(x − 2)| + C (b) log|(x − 2)/(x − 1)| + C (c) log|(x − 1)(x − 2)| + C (d) −log|(x − 1)(x − 2)| + C

Show the solution

The fraction is proper and the factors are distinct. Write 1/((x−1)(x−2)) = A/(x−1) + B/(x−2).
Multiply through: 1 = A(x − 2) + B(x − 1).
Put x = 1: 1 = A(−1), so A = −1.
Put x = 2: 1 = B(1), so B = 1.
Integrate: ∫[−1/(x−1) + 1/(x−2)] dx = −log|x−1| + log|x−2| + C.
Combine: log|x−2| − log|x−1| = log|(x−2)/(x−1)|.

Answer: (b) log|(x − 2)/(x − 1)| + C

Example 3

∫x·log x dx (log is natural log) equals: (a) (x²/2)·log x − x²/4 + C (b) (x²/2)·log x − x²/2 + C (c) (x²/2)·log x + x²/4 + C (d) x·log x − x²/4 + C

Show the solution

By ILATE, log x is Logarithmic and x is Algebraic. Logarithmic comes first, so u = log x and v = x.
u′ = 1/x and ∫v dx = x²/2.
Apply the formula: ∫x·log x dx = (x²/2)·log x − ∫(1/x)(x²/2) dx.
Simplify the new integral: ∫(x/2) dx = x²/4.
So the result is (x²/2)·log x − x²/4 + C.
Check by differentiating: d/dx[(x²/2)log x] = x·log x + x/2, and d/dx[x²/4] = x/2. The difference is x·log x, which matches.

Answer: (a) (x²/2)·log x − x²/4 + C

Exam tips

Questions are usually direct: one integral, four close options. Decide the method in a few seconds and do not over-explain to yourself.
Memorise the results for ∫log x dx, ∫x·eˣ dx and ∫x·log x dx. They save the most time.
Use differentiation to verify. Differentiating an option is often faster than finishing a long integral, and it catches sign errors.
Options often differ only in sign or in the order of the log ratio. Check A and B carefully before choosing.
If an integral needs repeated by parts or an unclear long division, skip it and return later. Negative marking is 0.25 per wrong answer.

Practice questions from Differential and Integral Calculus

Integration by Parts and Partial Fractions: frequently asked questions

What is the ILATE rule in integration by parts?

ILATE tells you which function to take as u. The order is Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. The function that comes earlier in this list is u, and the other function is v.

When should I use integration by parts and when partial fractions?

Use by parts when the integrand is a product of different types, such as x·eˣ or x·log x. Use partial fractions when the integrand is a rational function with a factorisable denominator. Try standard formulas or substitution first.

How do I integrate log x?

Write log x as 1·log x. Take u = log x and v = 1. This gives ∫log x dx = x·log x − x + C, where log is the natural log.

What if the fraction is not proper in partial fractions?

Divide the numerator by the denominator first. You get a polynomial quotient plus a proper fraction. Integrate the polynomial directly and split only the proper fraction into partial fractions.