Substitution, parts and partial fractions
Введено| English | Значение |
|---|---|
| integration by parts/ˌɪntɪˈɡreɪʃn baɪ pɑːts/ | интегрирование по частям |
Can we integrate the two factors separately?
- A simple-looking product such as xe^x cannot be integrated by separately integrating each factor.
- This lesson studies integration by parts 分部积分法: An integration method based on the derivative of a product.
The method
- Use substitution when an inner derivative appears as a factor.
- By parts, integral u v prime =uv-integral u prime v.
- For a rational function, divide first if needed, then decompose into partial fractions.
Why does the substitution w=x²+1 fit the integral of 2x/(x²+1)?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
With w=x²+1, dw=2x dx, and the integral becomes ∫dw/w=ln|w|+C. Substitution is powerful precisely when the inner derivative appears as a factor.
A worked case
For integral xe^x dx, take u=x and v prime=e^x. Then the integral is xe^x-e^x+C. For integral 2x/(x²+1) dx, substitute w=x²+1, dw=2x dx; the answer is ln(x²+1)+C.
Substitution, parts and partial fractions
integral from 0 to 3 of x² dx = 9
Compare signed area accumulation with the antiderivative x³/3; the substitution and parts examples are separate methods.
Evaluate the integral of xe^x from 0 to 1.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
An antiderivative is (x-1)e^x. At 1 and 0 it gives 0 and -1, so the integral is 1.
Where it goes wrong
- Choosing u and v prime well matters: the remaining integral should become simpler.
- In a definite substitution, either change the limits or return to x before applying the original limits.
Not true: The integral of a product equals the product of its separate integrals. A product needs a method such as integration by parts or substitution: ∫x·x dx=x³/3, not (x²/2)(x²/2).
Find the coefficient of ln(x²+1) in the integral of 2x/(x²+1).
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Substitution w=x²+1 gives dw=2x dx, so the coefficient of ln(x²+1) is 1.
The integral of a product equals the product of its separate integrals.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A product needs a method such as integration by parts or substitution: ∫x·x dx=x³/3, not (x²/2)(x²/2).
Applying it
- AQA H5 introduces substitution and parts; H6 integrates partial fractions with linear denominators.
- The single worked parts/substitution examples are introductory support, not complete repeated-parts or partial-fraction coverage.
Evaluate the integral of x² from 0 to 3.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
V/π=integral x² dx from 0 to 3=[x³/3]=9.
Match each integral to the method that fits it.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Choose the method by structure: an inner derivative suggests substitution, a product of unlike types suggests parts, and an improper rational function needs division first.
In your course
- AQA A-level Mathematics (7357), specification section H.
- Every paper allows a calculator, but method marks still need the expression or equation written before its value.
- Integration by parts: An integration method based on the derivative of a product.
- Use substitution when an inner derivative appears as a factor.
- Watch for: choosing u and v prime well matters: the remaining integral should become simpler.