Substitution, Parts, or Partial Fractions: Choose the Right Integration Method

The hardest part of many integration questions is not the calculation. It is choosing where to begin.

You may know substitution, integration by parts, and partial fractions separately, yet still pause when all three seem possible. That pause is normal. Integration is less like opening a formula box and more like reading clues at a crossroads.

The good news is that each method leaves a recognisable fingerprint:

• Substitution looks for an inner function and its derivative.
• Integration by parts looks for a product that becomes easier when one factor is differentiated.
• Partial fractions looks for a rational function that can be split into simpler fractions.

This guide will help you read that structure, test your choice quickly, and switch methods without wasting half a page.

First Ask: Can I Integrate It Directly?

Before choosing any special method, simplify the expression and check the standard forms you already know.

For example:

Integral of (3x^2 + 4cos x) dx

This needs no substitution, parts, or partial fractions. Integrate term by term:

= x^3 + 4sin x + C

Likewise, an expression may only need an identity or a small algebraic rearrangement. Factoring, dividing powers, separating terms, or completing a square can expose a standard integral immediately.

The Three Methods at a Glance

• Substitution: Look for a function with its derivative, or a constant multiple of it. Ask whether one replacement removes every x and reveals a standard form.
• Integration by parts: Look for a product with a factor that simplifies when differentiated. Ask whether the next integral will truly be easier.
• Partial fractions: Look for a ratio of polynomials with a factorable denominator. Ask whether splitting it will produce standard, simpler fractions.

The result matters more than the label. A method is not successful merely because you can start it. It is successful only if it makes the problem simpler.

A Reliable Decision Ladder

When a question does not yield directly, use this order.

Step 1: Simplify the integrand

Factor where useful. Divide polynomial powers. Apply a suitable identity. Check whether a numerator can be split.

Step 2: Look for the derivative of an inner expression

Circle a likely inner expression such as:

x^2 + 5
sin x
3x - 1
ln x

Differentiate it mentally. If that derivative, or a constant multiple of it, is also present, try substitution.

Step 3: If it is a rational function, inspect the denominator

A rational function has the form:

P(x) / Q(x)

where both P(x) and Q(x) are polynomials.

First compare degrees. If the numerator degree is at least the denominator degree, divide first. If the remaining fraction is proper and the denominator factors, partial fractions becomes a strong candidate.

But do not choose partial fractions automatically. If the numerator is already the derivative of the denominator, substitution may be much shorter.

Step 4: Look for a productive product

If the integrand is a product such as x e^x, x sin x, or x log x, ask whether differentiating one factor makes it simpler while the other factor is easy to integrate. If yes, use parts.

A lone logarithmic or inverse trigonometric function also often uses parts because you can treat it as multiplied by 1.

Step 5: Expect a chain of methods

Some integrals need substitution first and parts afterwards. Others need division before partial fractions. The first method is a doorway, not always the whole journey.

Method 1: Substitution Reverses the Chain Rule

Substitution is the natural choice when the integrand resembles:

f(g(x)) g'(x)

The inner expression is g(x), and its derivative g'(x) is also present.

The usual steps are:

1. Choose u = g(x).
2. Find du = g'(x) dx.
3. Rewrite the entire integral in terms of u.
4. Integrate.
5. Replace u with the original expression.

The most important check comes at Step 3: no stray x should remain.

Example 1: A clear inner-function pattern

Evaluate:

Integral of 2x / (x^2 + 5) dx

The denominator is the natural inner expression:

u = x^2 + 5
du = 2x dx

So:

Integral of 2x / (x^2 + 5) dx
= Integral of 1/u du
= ln|u| + C
= ln(x^2 + 5) + C

Since x^2 + 5 is always positive, the absolute-value signs are not needed in the final line.

Example 2: A product that is not a parts question

Evaluate:

Integral of x cos(x^2) dx

At first glance this is a product, but x is connected to the inner expression x^2.

Choose:

u = x^2
du = 2x dx
x dx = du/2

Then:

Integral of x cos(x^2) dx
= 1/2 Integral of cos u du
= 1/2 sin u + C
= 1/2 sin(x^2) + C

Using parts here would make a simple question much longer.

The constant-multiple test

The derivative need not appear exactly. A non-zero constant difference is harmless.

For example:

Integral of x / (3x^2 + 7) dx

The derivative of 3x^2 + 7 is 6x. The integrand contains x, which is one-sixth of 6x. Substitution still works.

What does not work is a missing variable factor. If you choose u = x^2 + 1 but the integrand has no x dx to create du, the substitution is probably incomplete or unsuitable.

Method 2: Integration by Parts Reverses the Product Rule

The formula is:

Int(u dv) = uv - Int(v du)

The goal is not simply to split a product. The goal is to exchange the original integral for an easier one.

A good choice usually has these two features:

• u becomes simpler when differentiated.
• dv can be integrated without difficulty.

The LIATE guideline can help you choose u:

Logarithmic
Inverse trigonometric
Algebraic
Trigonometric
Exponential

But LIATE is a starting clue, not a law. The decisive test is whether the remaining integral is easier.

Example 3: Algebraic times trigonometric

Evaluate:

Integral of (3x - 2)cos(2x) dx

Choose:

u = 3x - 2     dv = cos(2x) dx
du = 3 dx      v = 1/2 sin(2x)

Apply the formula:

Integral of (3x - 2)cos(2x) dx
= (3x - 2)/2 sin(2x) - 3/2 Integral of sin(2x) dx
= (3x - 2)/2 sin(2x) + 3/4 cos(2x) + C

The method works because differentiating 3x - 2 reduces it to the constant 3.

Example 4: A logarithm hiding a product

Evaluate:

Integral of x ln x dx

Choose the logarithm as u and place the algebraic factor with dv:

u = ln x       dv = x dx
du = 1/x dx    v = x^2/2

Then:

Integral of x ln x dx
= (x^2/2)ln x - Integral of (x^2/2)(1/x) dx
= (x^2/2)ln x - 1/2 Integral of x dx
= (x^2/2)ln x - x^2/4 + C

The remaining integral is simpler, so the choice is productive.

For a fuller treatment of choosing the two functions, see Integration by Parts: Choosing the First and Second Functions.

When parts is a poor first choice

Suppose you see:

Integral of x e^(x^2) dx

Parts is tempting because this is a product. But differentiating x and integrating e^(x^2) does not help because e^(x^2) has no elementary antiderivative by itself.

Substitution is immediate:

u = x^2
du = 2x dx

This is why the dv check is essential. Never choose a second function that you cannot integrate.

Method 3: Partial Fractions Simplifies a Rational Function

Partial fractions applies to a ratio of polynomials. It rewrites one complicated rational expression as a sum of simpler fractions.

Before decomposing, perform three checks.

Check 1: Is the expression rational?

Both numerator and denominator must be polynomials in x. An expression containing sin x, e^x, or ln x is not a rational function merely because it is written as a fraction.

Check 2: Is it proper?

For a proper rational function:

degree of numerator < degree of denominator

If this is false, use polynomial division first.

Check 3: Can the denominator be factored appropriately?

The decomposition depends on whether the denominator has distinct linear factors, repeated linear factors, or irreducible quadratic factors.

For distinct linear factors:

1 / [(x - a)(x - b)]
= A/(x - a) + B/(x - b)

For a repeated factor, include every power:

1 / [(x - a)^2(x - b)]
= A/(x - a) + B/(x - a)^2 + C/(x - b)

Leaving out one power makes the setup incomplete.

Example 5: Distinct linear factors

Evaluate:

Integral of (5x + 1) / (x^2 - x - 2) dx

Factor the denominator:

x^2 - x - 2 = (x - 2)(x + 1)

Set:

(5x + 1) / [(x - 2)(x + 1)]
= A/(x - 2) + B/(x + 1)

Multiply by (x - 2)(x + 1):

5x + 1 = A(x + 1) + B(x - 2)

Put x = 2:

11 = 3A
A = 11/3

Put x = -1:

-4 = -3B
B = 4/3

Therefore:

Integral of (5x + 1) / (x^2 - x - 2) dx
= 11/3 Integral of 1/(x - 2) dx
  + 4/3 Integral of 1/(x + 1) dx
= 11/3 ln|x - 2| + 4/3 ln|x + 1| + C

Recombining the two fractions is a quick way to verify the coefficients.

Example 6: Divide before decomposing

Evaluate:

Integral of (x^2 + 1) / (x^2 - x) dx

The numerator and denominator have the same degree, so divide first:

(x^2 + 1) / (x^2 - x)
= 1 + (x + 1)/(x^2 - x)
= 1 + (x + 1)/[x(x - 1)]

Now decompose the proper fraction:

(x + 1)/[x(x - 1)] = A/x + B/(x - 1)

So:

x + 1 = A(x - 1) + Bx

Putting x = 0 gives A = -1. Putting x = 1 gives B = 2.

Therefore:

Integral of (x^2 + 1) / (x^2 - x) dx
= Integral of [1 - 1/x + 2/(x - 1)] dx
= x - ln|x| + 2ln|x - 1| + C

Substitution or Partial Fractions? Use the Numerator Clue

A rational function can belong to either method. The numerator often makes the decision.

Consider:

Integral of (2x + 3) / (x^2 + 3x + 7) dx

The derivative of the denominator is exactly 2x + 3. Substitution solves it immediately:

u = x^2 + 3x + 7
du = (2x + 3) dx

So:

Integral of (2x + 3) / (x^2 + 3x + 7) dx
= ln(x^2 + 3x + 7) + C

There is no benefit in forcing partial fractions. In fact, the denominator does not factor into real linear factors.

Now compare:

Integral of (5x + 1) / (x^2 - x - 2) dx

The numerator is not a constant multiple of the denominator’s derivative 2x - 1, and the denominator factors neatly. Partial fractions is the cleaner choice.

Substitution or Parts? Use the Simplification Test

Compare these two integrals:

A. Integral of x cos(x^2) dx
B. Integral of x cos x dx

For A, the derivative of x^2 is present, so substitution removes the composition.

For B, there is no inner function whose derivative explains the other factor. But differentiating x simplifies it, while cos x is easy to integrate. Use parts.

The two expressions look similar. Their structures are different.

Integral	Structural clue	Method
Integral of x cos(x^2) dx	x matches the derivative of the inner x^2	Substitution
Integral of x cos x dx	Product with a reducible algebraic factor	Integration by parts

This comparison is worth remembering because it captures the whole decision process in two lines.

When One Question Needs Two Methods

Some questions deliberately test whether you can change methods midway.

Example 7: Substitution followed by parts

Evaluate:

Integral of x^3 e^(x^2) dx

The inner expression x^2 suggests substitution. Rewrite x^3 dx as x^2(x dx):

u = x^2
du = 2x dx
x dx = du/2

Then:

x^3 dx = x^2(x dx) = u du/2

So the integral becomes:

1/2 Integral of u e^u du

Now use integration by parts:

Integral of u e^u du = u e^u - e^u

Therefore:

Integral of x^3 e^(x^2) dx
= 1/2 e^u(u - 1) + C
= 1/2 e^(x^2)(x^2 - 1) + C

The first method exposed the second one.

A sensible rule for mixed questions

After each major step, pause and reclassify the new integral. Do not keep applying the original method out of habit.

Common Wrong Turns and Fast Corrections

Wrong turn 1: Calling every product a parts question

Correction: test whether one factor is the derivative of an inner function first.

x sin(x^2)   suggests substitution
x sin x      suggests parts

Wrong turn 2: Calling every fraction a partial-fractions question

Correction: partial fractions is for rational functions, and substitution may be shorter when the numerator matches the derivative of the denominator.

(2x + 1)/(x^2 + x + 4)
Use: substitution

1/[(x - 1)(x + 3)]
Use: partial fractions

Wrong turn 3: Choosing u but leaving stray x terms

If your substitution produces an integral containing both u and x, you have not completed the change of variable. Express every part in terms of u, or choose a different substitution.

Wrong turn 4: Skipping division

Do not decompose an improper rational function directly. Divide first, then decompose the proper remainder.

Wrong turn 5: Using LIATE without checking dv

The factor assigned to dv must be integrable. A perfect LIATE choice is useless if finding v is harder than the original problem.

Wrong turn 6: Forgetting absolute values

The standard result is:

Integral of 1/(x - a) dx = ln|x - a| + C

Keep the absolute-value signs unless the expression is known to stay positive on the domain being used.

Wrong turn 7: Trusting the method without checking the answer

Differentiate your final answer. This single habit catches missing constants, incorrect signs, incomplete decompositions, and algebra slips.

A 20-Second Method Check

When time is short, ask these questions in order:

1. Can I simplify or integrate directly?
2. Is there an inner expression whose derivative is present?
3. Is this a rational function?
4. If rational, is the numerator degree too large?
5. Does the denominator factor?
6. Is this a product where one factor simplifies on differentiation?
7. Can I integrate the proposed dv?
8. After the first step, has a new method appeared?

This sequence is more reliable than memorising a long list of isolated examples.

Practice Set: Name the Method Before Solving

For each integral, write the first method you would use.

1. Integral of 6x/(3x^2 + 4) dx
2. Integral of (3x - 2)cos(2x) dx
3. Integral of 1/[(x - 1)(x + 4)] dx
4. Integral of ln x dx
5. Integral of (x + 1)/(x^2 + 2x + 7) dx
6. Integral of (2x^2 + 1)/(x^2 - 1) dx
7. Integral of x^3 cos(x^2) dx
8. Integral of x sin(x^2 + 3) dx

Answers and reasons

1. Substitution. The numerator is the derivative of 3x^2 + 4.
2. Integration by parts. Differentiating the algebraic factor makes the remaining integral simpler.
3. Partial fractions. It is a proper rational function with distinct linear factors.
4. Integration by parts. Treat ln x as 1 times ln x.
5. Substitution. The numerator is half the derivative of the denominator.
6. Division, then partial fractions. The rational function is improper.
7. Substitution, then integration by parts. Let u = x^2; the new integral contains u cos u.
8. Substitution. Let u = x^2 + 3.

How to Build Method Recognition

Do not practise all questions with the method already written above them. That trains calculation but not selection.

Instead, make a mixed page of integrals and use three passes:

1. Classify only. Write direct, substitution, parts, partial fractions, or mixed beside each question.
2. Write the clue. Note one reason, such as inner derivative present or proper rational function with factorable denominator.
3. Solve and differentiate. Complete the calculation, then verify the result.

Ten minutes of mixed classification can improve decision-making more than another hour of doing questions grouped under obvious headings.

Sources and Further Reading

• NCERT Mathematics Part II: Integrals, including the chapter’s standard forms and worked methods.
• OpenStax Calculus Volume 1: Substitution, which explains substitution as reversing a chain-rule pattern.
• OpenStax Calculus Volume 2: Integration by Parts, including method choice and repeated use.
• OpenStax Calculus Volume 2: Partial Fractions, including division, factor types, and coefficient checks.
• MIT OpenCourseWare integration techniques notes, with practice across substitution, parts, and partial fractions.

Frequently Asked Questions

How do I decide the integration method quickly?

Simplify first. Then look for an inner function with its derivative, which suggests substitution. If the expression is rational, check division and factorisation for partial fractions. If it is a productive product, test integration by parts.

Does every product require integration by parts?

No. A product such as x cos(x^2) is better handled by substitution because x matches the derivative of the inner expression x^2. Use parts only when differentiating one factor and integrating the other creates an easier integral.

Does every rational function require partial fractions?

No. If the numerator is the derivative of the denominator, substitution may solve it immediately. Partial fractions is most useful when a proper rational function has a denominator that can be decomposed into suitable factors.

What should I do if the numerator degree is greater than or equal to the denominator degree?

Use polynomial division first. Then apply partial fractions only to the proper rational remainder.

How do I know whether my substitution is complete?

After replacing the inner expression and its differential, the integral should contain only the new variable. If both x and u remain, rewrite the leftover terms or choose a better substitution.

Is LIATE always correct for integration by parts?

No. LIATE is a useful guide for choosing u, but you must also check that dv is easy to integrate and that the new integral is simpler.

Can one integral need more than one method?

Yes. An integral may require substitution followed by parts, or polynomial division followed by partial fractions. Reclassify the integral after every major step.

Why do logarithms from partial fractions need absolute-value signs?

Because ln|x - a| differentiates to 1/(x - a) wherever the expression is defined, whether x - a is positive or negative.

How can I check a partial-fraction decomposition?

Combine the separated fractions again. If their numerator and denominator reproduce the original rational function, the decomposition is correct.

What is the fastest way to check my final integral?

Differentiate the answer. If you recover the original integrand, your result is correct apart from the arbitrary constant.

Should I memorise a separate rule for every integral?

No. Memorise the structural clues and standard antiderivatives. The same decision process then works across many different-looking questions.

What should I practise if I know the methods but choose the wrong one?

Practise mixed sets without method headings. Classify each question and write one reason before solving. That directly trains the skill you are missing.