Integration as reverse differentiation and standard integrals
If the speedometer readings are known at every instant, the distance travelled is found by adding up tiny pieces; reversing the derivative does exactly this. Building on the previous chapter : every derivative formula can be read backwards.
If F′(x) = f(x), then ∫ f(x) dx = F(x) + C; F is an antiderivative and C is the constant of integration. Linearity: ∫(af + bg) dx = a∫f dx + b∫g dx.
Rule 1: Powers — ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1); ∫(1/x) dx = log|x| + C; ∫eˣ dx = eˣ + C.
Rule 2: Trigonometric — ∫sin x dx = −cos x + C; ∫cos x dx = sin x + C; ∫sec²x dx = tan x + C.
Rule 3: Special forms — ∫dx/(x² + a²) = (1/a)tan⁻¹(x/a) + C; ∫dx/√(a² − x²) = sin⁻¹(x/a) + C.
How is the formula obtained? : ∫(3x² + 2x + 1) dx = 3·x³/3 + 2·x²/2 + x + C = x³ + x² + x + C. Differentiate to check: 3x² + 2x + 1.
- Easy : ∫(4x³ − 2) dx = x⁴ − 2x + C.
- Medium : ∫sin 2x dx = −(cos 2x)/2 + C.
- Exam-type : ∫dx/(x² + 9) = (1/3)tan⁻¹(x/3) + C; ∫dx/√(4 − x²) = sin⁻¹(x/2) + C.
Always add + C; check by differentiating the answer.
Forgetting the factor 1/a for ∫f(ax + b) dx, which equals F(ax + b)/a.
For ∫f(ax + b) dx divide the standard result by a. Memory clue : "Integrate, divide by the inside slope, add C."
∫(1/x) dx = log|x|; the absolute value matters for negative x.
CET-style example : ∫ dx/(x² + 9) = (A) (1/3)tan⁻¹(x/3) + C (B) 3tan⁻¹(x/3) + C (C) tan⁻¹(3x) + C (D) log(x² + 9) + C
Answer : (A). Time-saving method : Use the a² form with a = 3.
Spot the difference
| Integral | Result | ||
|---|---|---|---|
| ∫xⁿ dx | xⁿ⁺¹/(n + 1) | ||
| ∫1/x dx | log | x | |
| ∫sin(ax) dx | −cos(ax)/a |
Practice questions
- 1∫x² dx
- 2∫(3x² + 2x + 1) dx
- 3∫cos x dx
- 4∫sin 2x dx
- 5∫e^{3x} dx
- 6∫dx/(x² + 9)
- 7∫dx/√(4 − x²)
- 8∫(x + 1/x)² dx
Show answers (try them yourself first)
- 1x³/3 + C.
- 2x³ + x² + x + C.
- 3sin x + C.
- 4−cos 2x/2 + C.
- 5e^{3x}/3 + C.
- 6(1/3)tan⁻¹(x/3) + C.
- 7sin⁻¹(x/2) + C.
- 8∫(x² + 2 + 1/x²) dx = x³/3 + 2x − 1/x + C.
Integration by substitution
Converting kilometres to metres before adding distances makes the sum easy; substitution changes the variable so that the integral becomes a standard one.
If the integrand contains a function and its derivative, put u = g(x), du = g′(x) dx and integrate in u. Then replace u by g(x). Useful pattern: ∫f′(x)/f(x) dx = log|f(x)| + C.
Rule 1: Choose u — Let u be the inside function whose derivative (up to a constant) is also present.
Rule 2: Back-substitute — Return to x at the end and add C.
How is the formula obtained? : ∫x e^{x²} dx: let u = x², du = 2x dx, so x dx = du/2. ∫e^u (du/2) = e^u/2 + C = e^{x²}/2 + C.
- Easy : ∫2x/(x² + 1) dx = log(x² + 1) + C (u = x² + 1).
- Medium : ∫x e^{x²} dx = e^{x²}/2 + C.
- Exam-type : ∫sin²x cos x dx = sin³x/3 + C (u = sin x).
A constant factor missing from du is fixed by multiplying and dividing.
Leaving the answer in u.
If the numerator is the derivative of the denominator the answer is log of the denominator. Memory clue : "Inside function u, its derivative du."
If the derivative of u is not present after adjusting a constant, substitution is the wrong tool.
CET-style example : ∫2x/(x² + 1) dx = (A) log(x² + 1) + C (B) 2 log x + C (C) tan⁻¹x + C (D) (x² + 1)² + C
Answer : (A). Time-saving method : Numerator is the derivative of the denominator.
Spot the difference
| Pattern | Answer | ||
|---|---|---|---|
| ∫f′/f dx | log | f | + C |
| ∫f(x)ⁿ f′(x) dx | f(x)ⁿ⁺¹/(n + 1) + C |
Practice questions
- 1∫2x/(x² + 1) dx
- 2∫(2x + 3)⁴ dx
- 3∫x e^{x²} dx
- 4∫cos x/sin x dx
- 5∫sin²x cos x dx
- 6∫x/(x² + 4) dx
- 7∫tan x dx
- 8∫e^x/(1 + e^x) dx
Show answers (try them yourself first)
- 1log(x² + 1) + C.
- 2(2x + 3)⁵/10 + C.
- 3e^{x²}/2 + C.
- 4log|sin x| + C.
- 5sin³x/3 + C.
- 6(1/2)log(x² + 4) + C.
- 7−log|cos x| + C = log|sec x| + C.
- 8log(1 + e^x) + C.
Integration by parts
To find the total of a product of two changing things (like price × quantity that both vary) we trade the hard part for an easy one, the way a seesaw swaps load.
∫u v dx = u ∫v dx − ∫[u′ ∫v dx] dx. Choose u by the order ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential): the earlier letter is u.
Rule 1: Formula — ∫u dv = uv − ∫v du.
Rule 2: Repeated — If the new integral is again a product, apply the rule once more.
How is the formula obtained? : ∫x eˣ dx: u = x, v = eˣ ⇒ x eˣ − ∫eˣ dx = x eˣ − eˣ + C = (x − 1)eˣ + C.
- Easy : ∫x eˣ dx = (x − 1)eˣ + C.
- Medium : ∫log x dx = x log x − x + C (u = log x, v = 1).
- Exam-type : ∫x log x dx = (x²/2)log x − x²/4 + C.
ILATE tells you which factor to differentiate.
Choosing u = eˣ and v = x, which makes the integral harder.
If one factor is log or inverse trig, take it as u. Memory clue : "ILATE: the earlier letter is u."
Differentiate u, integrate v; do not mix them up.
CET-style example : ∫x eˣ dx = (A) (x − 1)eˣ + C (B) (x + 1)eˣ + C (C) x eˣ + C (D) x²eˣ/2 + C
Answer : (A). Time-saving method : Differentiate the answer.
Spot the difference
| Choose u as | Because |
|---|---|
| log x | its derivative 1/x is simpler |
| x (algebraic) | its derivative is a constant |
| eˣ | only as v; it never gets simpler |
Practice questions
- 1∫x eˣ dx
- 2∫log x dx
- 3∫x log x dx
- 4∫x cos x dx
- 5∫x sin x dx
- 6∫tan⁻¹x dx
- 7∫x² eˣ dx
- 8∫eˣ(sin x + cos x) dx
Show answers (try them yourself first)
- 1(x − 1)eˣ + C.
- 2x log x − x + C.
- 3(x²/2)log x − x²/4 + C.
- 4x sin x + cos x + C.
- 5−x cos x + sin x + C.
- 6x tan⁻¹x − (1/2)log(1 + x²) + C.
- 7eˣ(x² − 2x + 2) + C.
- 8eˣ sin x + C.
Integration by partial fractions
A mixed juice can be separated into its simple ingredients; a complicated fraction is split into simple fractions that are easy to integrate.
A proper rational function with factorised denominator is written as a sum of simple fractions: 1/((x + a)(x + b)) = A/(x + a) + B/(x + b). Find A and B by comparing, or by putting x equal to each root.
Rule 1: Split — Linear factors give A/(x + a); a repeated factor gives A/(x + a) + B/(x + a)².
Rule 2: Integrate — ∫dx/(x + a) = log|x + a| + C.
How is the formula obtained? : x/((x + 1)(x + 3)) = A/(x + 1) + B/(x + 3). Put x = −1: A = (−1)/2 = −1/2. Put x = −3: B = (−3)/(−2) = 3/2. Integral = −½ log|x + 1| + (3/2) log|x + 3| + C.
- Easy : 1/((x + 1)(x + 2)) = 1/(x + 1) − 1/(x + 2) ⇒ ∫ = log|(x + 1)/(x + 2)| + C.
- Medium : 2/(x² − 1) = 1/(x − 1) − 1/(x + 1) ⇒ ∫ = log|(x − 1)/(x + 1)| + C.
- Exam-type : ∫x/((x + 1)(x + 3)) dx = −½ log|x + 1| + (3/2) log|x + 3| + C.
Check A + B or a value at x = 0 to confirm the split.
Splitting an improper fraction without dividing first.
Cover-up method: cover one factor and put its root in the rest. Memory clue : "Cover the factor, put in the root."
If degree of numerator ≥ degree of denominator, divide first.
CET-style example : 1/((x + 1)(x + 2)) = (A) 1/(x + 1) − 1/(x + 2) (B) 1/(x + 1) + 1/(x + 2) (C) 2/(x + 1) − 1/(x + 2) (D) 1/(x + 2) − 1/(x + 1)
Answer : (A). Time-saving method : Cover-up at x = −1 and x = −2.
Spot the difference
| Denominator | Split form |
|---|---|
| (x + a)(x + b) | A/(x + a) + B/(x + b) |
| (x + a)² | A/(x + a) + B/(x + a)² |
Practice questions
- 1Split 1/((x + 1)(x + 2)).
- 2∫dx/((x + 1)(x + 2))
- 3Split 2/(x² − 1).
- 4∫2/(x² − 1) dx
- 5Split x/((x + 1)(x + 3)).
- 6∫x/((x + 1)(x + 3)) dx
- 7∫dx/(x² − 4)
- 8∫(3x + 5)/((x + 1)(x + 2)) dx
Show answers (try them yourself first)
- 11/(x + 1) − 1/(x + 2).
- 2log|(x + 1)/(x + 2)| + C.
- 31/(x − 1) − 1/(x + 1).
- 4log|(x − 1)/(x + 1)| + C.
- 5−½/(x + 1) + (3/2)/(x + 3).
- 6−½ log|x + 1| + (3/2) log|x + 3| + C.
- 7(1/4) log|(x − 2)/(x + 2)| + C.
- 8A = 2, B = 1 ⇒ 2 log|x + 1| + log|x + 2| + C.