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Integration methods

Integration by Substitution: Match the Inner Derivative

Substitution reverses the chain rule. It works when one part is a composite input and another part supplies its derivative, possibly up to a constant.

Direct answer

If u=g(x), then ∫f(g(x))g′(x)dx=∫f(u)du

The substitution must account for the differential factor and preserve or transform the bounds.

When to use integration by substitution

Choose the method from the top-level mathematical structure before manipulating symbols. Conditions are part of the task, not optional notes.

Ask first 1

Which structure controls the method?

What to do

Prefer substitution when an inner expression and its derivative factor are both visible.

Ask first 2

Which condition can change the answer?

What to do

The substitution must account for the differential factor and preserve or transform the bounds.

Ask first 3

How should the result be checked?

What to do

Differentiate the answer and confirm every original factor returns.

Reverse the chain rule

Let F′(u)=f(u) and set u=g(x).

The chain rule gives d/dx F(g(x))=f(g(x))g′(x).

Therefore the antiderivative is F(g(x))+C.

For definite integrals, transform both bounds or return to x before evaluating; never mix variables and bounds.

1
u=g xu = g\,x
2
du=g'(x)dx
3
∫f(g(x))g'(x)dx=F(g(x))+C

Worked examples

∫2x cos(x²) dx
sin⁡(x2)+C\sin\left(x^{2}\right) + C

Use u=x².

∫3 x2x3+1 dx\int \frac{3\,x^{2}}{x^{3} + 1}\,\mathrm{d}x
ln⁡∣x3+1∣+C\ln\left|x^{3} + 1\right| + C

Use u=x³+1.

∫₀¹ 2x exp(x²) dx
e−1e - 1

Transform bounds to u=0 and u=1.

Common mistakes

  • • Leaving unmatched x terms.
  • • Forgetting a constant adjustment.
  • • Mixing x-bounds with u.
  • • Not differentiating the result.

Error clinic: locate the first invalid step

Incorrect attempt

∫2x cos(x²)dx=sin(x)+C

Why it fails

The input x² was not restored.

Correction

sin⁡(x2)+C\sin\left(x^{2}\right) + C

Incorrect attempt

u=x³+1, du=x²dx

Why it fails

The factor 3 is missing.

Correction

du=3x²dx

Practice the same idea with a changed structure

Name the rule and its conditions before revealing the reference answer. Explain every sign and factor.

∫3 x2x3+1 dx\int \frac{3\,x^{2}}{x^{3} + 1}\,\mathrm{d}x

Check before solving

Use u=x³+1.

Reveal the reference answer

Reference answer

ln⁡∣x3+1∣+C\ln\left|x^{3} + 1\right| + C
∫₀¹ 2x exp(x²) dx

Check before solving

Transform bounds to u=0 and u=1.

Reveal the reference answer

Reference answer

e−1e - 1

Try the worked examples in the calculator

Each link opens the matching calculator with the expression, variable and required conditions already filled in.

Integrais:2 x cos⁡(x2)2\,x\,\cos\left(x^{2}\right)Integrais:3 x2x3+1\frac{3\,x^{2}}{x^{3} + 1}
Open the calculator without a preset: Integrais

Frequently asked questions

How do I choose u?+

Look for an inner expression whose derivative appears elsewhere.

Can du differ by a constant?+

Yes. Adjust the constant explicitly.

Must bounds change?+

Change them if you remain in u; otherwise substitute back before using x-bounds.

Conclusion

Choose a substitution that absorbs both the composite input and its differential, then verify by differentiation.

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