Intuition The ONE big ideaIntegration is just anti-differentiation : "what function did I differentiate to get this?"
Every rule below is a derivative rule read backwards . If you know your derivatives cold,
you already know all your integrals — you just have to read the table from right to left, and
tack on a + C +C + C destroys constants.
Intuition The lost-constant problemd d x ( x 2 ) = 2 x \frac{d}{dx}(x^2) = 2x d x d ( x 2 ) = 2 x d d x ( x 2 + 7 ) = 2 x \frac{d}{dx}(x^2 + 7) = 2x d x d ( x 2 + 7 ) = 2 x d d x ( x 2 − 99 ) = 2 x \frac{d}{dx}(x^2 - 99) = 2x d x d ( x 2 − 99 ) = 2 x flattens any constant to 0 0 0 + C +C + C
Definition Indefinite integral (antiderivative)∫ f ( x ) d x = F ( x ) + C means F ′ ( x ) = f ( x ) . \displaystyle\int f(x)\,dx = F(x) + C \quad\text{means}\quad F'(x) = f(x). ∫ f ( x ) d x = F ( x ) + C means F ′ ( x ) = f ( x ) . ∫ \int ∫ sum ), and d x dx d x variable of integration .
We want a function whose derivative is x n x^n x n go up we should raise the power by 1 — and then fix the leftover constant.
Derivation. Guess that the antiderivative looks like x n + 1 x^{n+1} x n + 1 d d x x n + 1 = ( n + 1 ) x n . \frac{d}{dx}\,x^{n+1} = (n+1)\,x^{n}. d x d x n + 1 = ( n + 1 ) x n . ( n + 1 ) (n+1) ( n + 1 ) Why this step? Differentiation multiplied by the old power+1, so
we cancel it by dividing by ( n + 1 ) (n+1) ( n + 1 ) d d x [ x n + 1 n + 1 ] = ( n + 1 ) x n n + 1 = x n . ✓ \frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = \frac{(n+1)x^n}{n+1} = x^n. \checkmark d x d [ n + 1 x n + 1 ] = n + 1 ( n + 1 ) x n = x n . ✓
n = − 1 n=-1 n = − 1 n = − 1 n=-1 n = − 1 feels right because the
rule works for every other number. But at n = − 1 n=-1 n = − 1 n + 1 = 0 n+1 = 0 n + 1 = 0 ∫ x − 1 d x = ln ∣ x ∣ + C \int x^{-1}dx = \ln|x| + C ∫ x − 1 d x = ln ∣ x ∣ + C
Intuition HOW to get these for freeYou memorised d d x sin x = cos x \frac{d}{dx}\sin x = \cos x d x d sin x = cos x d d x cos x = − sin x \frac{d}{dx}\cos x = -\sin x d x d cos x = − sin x
Derivation of ∫ cos x d x \int \cos x\,dx ∫ cos x d x We want F F F F ′ = cos x F' = \cos x F ′ = cos x d d x sin x = cos x \frac{d}{dx}\sin x=\cos x d x d sin x = cos x F = sin x F=\sin x F = sin x
Derivation of ∫ sin x d x \int \sin x\,dx ∫ sin x d x We want F ′ = sin x F' = \sin x F ′ = sin x d d x cos x = − sin x \frac{d}{dx}\cos x = -\sin x d x d cos x = − sin x negative of what we want. Why this step? Multiply by − 1 -1 − 1 d d x ( − cos x ) = sin x . \frac{d}{dx}(-\cos x) = \sin x. d x d ( − cos x ) = sin x . ∫ sin x d x = − cos x + C \int\sin x\,dx = -\cos x + C ∫ sin x d x = − cos x + C
Students write ∫ sin x d x = cos x \int\sin x\,dx = \cos x ∫ sin x d x = cos x feels symmetric with the cosine rule. The fix:
derivatives of sin/cos carry a minus sign on the cosine side , so it shows up in BOTH integrals.
Quick self-check: differentiate your answer — it must reproduce the integrand.
e x e^x e x e x e^x e x fixed point of differentiation: d d x e x = e x \frac{d}{dx}e^x = e^x d x d e x = e x
∫ e x d x = e x + C . \int e^x\,dx = e^x + C. ∫ e x d x = e x + C .
For a general base a > 0 a>0 a > 0 d d x a x = a x ln a \frac{d}{dx}a^x = a^x\ln a d x d a x = a x ln a ln a \ln a ln a ln a \ln a ln a ∫ a x d x = a x ln a + C . \int a^x\,dx = \frac{a^x}{\ln a} + C. ∫ a x d x = l n a a x + C .
Intuition WHAT fills the holeWe need a function whose derivative is 1 x \frac1x x 1 d d x ln x = 1 x \frac{d}{dx}\ln x = \frac1x d x d ln x = x 1 ln x \ln x ln x 1 x \frac1x x 1 x x x ln x \ln x ln x ln ∣ x ∣ \ln|x| ln ∣ x ∣ d d x ln ∣ x ∣ = 1 x \frac{d}{dx}\ln|x| = \frac1x d x d ln ∣ x ∣ = x 1 all x ≠ 0 x\neq 0 x = 0
∫ 1 x d x = ln ∣ x ∣ + C . \int \frac{1}{x}\,dx = \ln|x| + C. ∫ x 1 d x = ln ∣ x ∣ + C .
Intuition WHY integrals splitDifferentiation is linear: d d x [ a f + b g ] = a f ′ + b g ′ \frac{d}{dx}[af+bg] = af'+bg' d x d [ a f + b g ] = a f ′ + b g ′
Worked example 1 — Power rule with a fraction∫ x d x \displaystyle\int \sqrt{x}\,dx ∫ x d x
Rewrite x = x 1 / 2 \sqrt x = x^{1/2} x = x 1/2 Why? The rule needs a power, not a root.
n = 1 2 n=\tfrac12 n = 2 1 n + 1 = 3 2 n+1=\tfrac32 n + 1 = 2 3 x 3 / 2 3 / 2 + C = 2 3 x 3 / 2 + C \dfrac{x^{3/2}}{3/2}+C = \dfrac{2}{3}x^{3/2}+C 3/2 x 3/2 + C = 3 2 x 3/2 + C Check: d d x 2 3 x 3 / 2 = 2 3 ⋅ 3 2 x 1 / 2 = x 1 / 2 \frac{d}{dx}\frac23x^{3/2} = \frac23\cdot\frac32 x^{1/2}=x^{1/2} d x d 3 2 x 3/2 = 3 2 ⋅ 2 3 x 1/2 = x 1/2
1 / x 1/x 1/ x ∫ 6 x d x \displaystyle\int \frac{6}{x}\,dx ∫ x 6 d x
Why not power rule? Here n = − 1 n=-1 n = − 1 6 ∫ 1 x d x 6\int\frac1x dx 6 ∫ x 1 d x = 6 ln ∣ x ∣ + C = 6\ln|x| + C = 6 ln ∣ x ∣ + C
Worked example 4 — Mixed trig & exponential∫ ( 2 e x − sin x + sec 2 x ) d x . \displaystyle\int (2e^x - \sin x + \sec^2 x)\,dx. ∫ ( 2 e x − sin x + sec 2 x ) d x .
∫ 2 e x = 2 e x \int 2e^x = 2e^x ∫ 2 e x = 2 e x ∫ − sin x = + cos x \int -\sin x = +\cos x ∫ − sin x = + cos x ∫ sec 2 x = tan x \int \sec^2 x = \tan x ∫ sec 2 x = tan x Answer: 2 e x + cos x + tan x + C 2e^x + \cos x + \tan x + C 2 e x + cos x + tan x + C
Worked example 5 — General base∫ 2 x d x = 2 x ln 2 + C . \displaystyle\int 2^x\,dx = \dfrac{2^x}{\ln 2} + C. ∫ 2 x d x = ln 2 2 x + C . d d x 2 x ln 2 = 2 x ln 2 ln 2 = 2 x . \frac{d}{dx}\frac{2^x}{\ln 2}=\frac{2^x\ln 2}{\ln 2}=2^x. d x d l n 2 2 x = l n 2 2 x l n 2 = 2 x .
Recall Feynman: explain to a 12-year-oldDifferentiating is like asking "how fast is this changing?" Integrating is the detective
version: I tell you the speed at every moment and you reconstruct where the car was. But there's
a catch — you don't know the car's starting point , so every answer ends with "+ some unknown
start" (that's the + C +C + C
Mnemonic Remembering the rules"Power UP & divide, exp stays alive, sin→minus-cos drives, and one-over-x logs survives."
(Raise the power then divide; e x e^x e x ∫ sin = − cos \int\sin=-\cos ∫ sin = − cos ∫ 1 x = ln ∣ x ∣ \int\frac1x=\ln|x| ∫ x 1 = ln ∣ x ∣
Cover the answers and predict each, then differentiate to verify (Forecast-then-Verify).
What does + C +C + C The lost constant — differentiation sends any constant to 0, so the antiderivative can't recover it;
+ C +C + C encodes all possibilities.
State the power rule and its one exception. ∫ x n d x = x n + 1 n + 1 + C \int x^n dx = \frac{x^{n+1}}{n+1}+C ∫ x n d x = n + 1 x n + 1 + C for
n ≠ − 1 n\neq -1 n = − 1 ; at
n = − 1 n=-1 n = − 1 you'd divide by zero.
∫ 1 x d x = ? \int \frac1x dx = ? ∫ x 1 d x = ? ∣ x ∣ |x| ∣ x ∣ ln ∣ x ∣ + C \ln|x|+C ln ∣ x ∣ + C ; absolute value extends validity to
x < 0 x<0 x < 0 since
d d x ln ∣ x ∣ = 1 x \frac{d}{dx}\ln|x|=\frac1x d x d ln ∣ x ∣ = x 1 everywhere except 0.
∫ sin x d x = ? \int \sin x\,dx = ? ∫ sin x d x = ? − cos x + C -\cos x + C − cos x + C (note the minus sign).
∫ cos x d x = ? \int \cos x\,dx = ? ∫ cos x d x = ? ∫ sec 2 x d x = ? \int \sec^2 x\,dx = ? ∫ sec 2 x d x = ? ∫ e x d x = ? \int e^x dx = ? ∫ e x d x = ? e x + C e^x + C e x + C ;
e x e^x e x is the fixed point of differentiation.
∫ a x d x = ? \int a^x dx = ? ∫ a x d x = ? a x ln a + C \frac{a^x}{\ln a}+C l n a a x + C , derived by dividing out the extra
ln a \ln a ln a from
d d x a x \frac{d}{dx}a^x d x d a x .
Is there a product rule for integration? No — products need substitution or integration by parts.
How do you always verify an integral? Differentiate your answer; it must reproduce the integrand.
Integration = anti-differentiation
Indefinite integral F plus C
Integral of 1 over x = ln abs x
Intuition Hinglish mein samjhoDekho, integration ka asli matlab hai ulta differentiation — "aisa kaunsa function tha jise
differentiate karke ye mila?" Bas. Tum already saari derivatives jaante ho, to bas table ko
right-to-left padho. Power rule yaad rakhna easy hai: power ko ek badhao aur usi se divide karo ,
yani ∫ x n d x = x n + 1 n + 1 + C \int x^n dx = \frac{x^{n+1}}{n+1}+C ∫ x n d x = n + 1 x n + 1 + C n = − 1 n=-1 n = − 1 0 0 0 ln ∣ x ∣ \ln|x| ln ∣ x ∣
Trig wala part muft milta hai agar derivatives pakke hain. d d x sin x = cos x \frac{d}{dx}\sin x=\cos x d x d sin x = cos x ∫ cos x = sin x \int\cos x=\sin x ∫ cos x = sin x d d x cos x = − sin x \frac{d}{dx}\cos x=-\sin x d x d cos x = − sin x ∫ sin x = − cos x \int\sin x = -\cos x ∫ sin x = − cos x minus sign
mat bhoolna , yahi sabse common galti hai. Exponential mein e x e^x e x ln a \ln a ln a
Ye + C +C + C 0 0 0 Forecast-then-Verify se exam mein kabhi galti nahi hogi.
