Basic integration rules — power, trig, exponential, log: Worked examples

4.2.2 · D3 · Maths › Calculus II — Integration › Basic integration rules — power, trig, exponential, log

Hume sirf anti-differentiation ("main kya differentiate karke yeh result laaya?") aur linearity (sums ko split karo, constants bahar nikalo) chahiye. Neeche har symbol parent mein earn kiya gaya hai; jahan koi naya idea aata hai, main use yahan rebuild karta hoon.

Scenario matrix

Har basic integral ko is table ke ek cell mein landing karte socho. Agar hum ek cell per ek worked example cover kar lein, toh hum sab kuch cover kar lete hain jo yeh topic throw kar sakta hai.

Cell	Case class	Kya tricky hai	Kis Example mein aaya
A	Positive integer power	plain power rule	1
B	Fractional / root power	pehle root ko exponent mein rewrite karo	1
C	Negative power ( $n \leq - 2$ )	$n + 1$ ka sign flip ho jaata hai	2
D	The forbidden $n = - 1$	power rule kaam nahi karta, log le leta hai	3
E	Trig with the minus-sign trap	$\int sin = - cos$	4
F	Exponential, base $e$ vs base $a$	sirf base $a$ ke liye $ln a$ se divide karo	5
G	Zero / degenerate input	integrand $= 0$ , ya ek constant	6
H	Real-world word problem	speed → position, $+ C$ choose karo	7
I	Exam twist: hidden power	integrate karne se pehle algebra karo	8
J	Limiting/edge behaviour	kya hota hai jab $n \to - 1$	9

Neeche, har example un cells ka naam leta hai jinhein woh clear karta hai.

Neeche wala figure isi matrix ko ek coloured grid ki tarah draw karta hai — har pastel block ek cell hai, aur colour sirf related cases ko group karta hai (purple = power-type, coral = tricky negative/log region, mint = trig & exponential, butter = degenerate/word-problem). Isse checklist ki tarah use karo: har block exactly ek worked example se clear hota hai.

Figure — Basic integration rules — power, trig, exponential, log

Worked examples

Worked example Example 1 — Cells A & B: integer aur root powers saath mein

$\int (x^{3} + 3 x) d x .$

Forecast: har power ko ek badhao aur divide karo. Guess karo ki cube root kuch aisa banega jaise $x^{4/3}$ — lekin aage kaunsa fraction hoga?

Root ko power ki tarah rewrite karo. $3 x = x^{1/3}$ . Yeh step kyun? Power rule sirf $x^{n}$ form ko pehchaanta hai; root symbol uske liye invisible hai. $3 x$ ko $x^{1/3}$ mein convert karne se $n = \frac{1}{3}$ visible ho jaata hai. Domain note. Real cube root $3 x$ saare $x$ ke liye defined hai (negatives bhi), lekin fractional-power symbol $x^{1/3}$ sirf $x \geq 0$ ke liye unambiguous hai. Toh hum rewrite — aur final answer — $x \geq 0$ domain par karte hain; $x < 0$ ke liye $x^{1/3}$ ko real cube root samjho aur formula tab bhi correctly differentiate hoga.
$x^{3}$ par power rule apply karo. Yahan $n = 3$ , toh $n + 1 = 4$ : $\frac{x ^{4}}{4}$ . Yeh step kyun? Differentiation power ko ek se giraata hai, toh isse ulta chalane ke liye power badhao aur naye power se divide karo taaki leftover factor cancel ho jaye.
$x^{1/3}$ par power rule apply karo. Yahan $n + 1 = \frac{1}{3} + 1 = \frac{4}{3}$ : $\frac{x ^{4/3}}{4/3} = \frac{3}{4} x^{4/3}$ . Yeh step kyun? Fraction $\frac{4}{3}$ se divide karna matlab uske reciprocal $\frac{3}{4}$ se multiply karna — woh reciprocal hi hai jisse aage $\frac{3}{4}$ aata hai.
$+ C$ add karo. $\frac{x ^{4}}{4} + \frac{3}{4} x^{4/3} + C .$

Verify karo: differentiate karo. $\frac{d}{d x} \frac{x ^{4}}{4} = x^{3}$ ✓, aur $\frac{d}{d x} \frac{3}{4} x^{4/3} = \frac{3}{4} \cdot \frac{4}{3} x^{1/3} = x^{1/3} = 3 x$ ✓.

Worked example Example 3 — Cell D: the forbidden

$n = - 1$ $\int \frac{- 2}{x} d x .$

Forecast: yeh $n = - 1$ wali power jaisi lagti hai. Lekin power rule $n + 1 = 0$ se divide karega. Kya iska matlab hai integral exist hi nahi karta? (Nahi — ek alag function hume bachata hai.)

$n = - 1$ pehchano. $\frac{- 2}{x} = - 2 x^{- 1}$ , toh $n = - 1$ aur $n + 1 = 0$ . Yeh step kyun? $n = - 1$ spot karna hi poora game hai — yeh woh ek value hai jo power rule forbid karta hai kyunki zero se divide nahi ho sakta.
Log rule par switch karo. $\int \frac{1}{x} d x = ln ∣ x ∣ + C$ (yahi woh function hai jo gap fill karta hai). Yeh step kyun? Hume ek aisa function chahiye jiska derivative $\frac{1}{x}$ ho. Natural log and exponential functions se, $\frac{d}{d x} ln ∣ x ∣ = \frac{1}{x}$ saare $x \neq = 0$ ke liye — negative $x$ ke liye bhi, isliye absolute value $∣ x ∣$ wahan hai.
Constant bahar nikalo. $- 2 ln ∣ x ∣ + C .$

Verify karo: $\frac{d}{d x} (- 2 ln ∣ x ∣) = - 2 \cdot \frac{1}{x} = \frac{- 2}{x}$ ✓, har $x \neq = 0$ ke liye valid.

Worked example Example 4 — Cell E: sin/cos minus-sign trap

$\int (4 cos x - 3 sin x) d x .$

Forecast: in mein se ek integrate hone par minus sign pick up karta hai, aur yeh woh nahi jo tum expect karte ho. Kaunsa term flip hoga?

$4 cos x$ integrate karo. Kyunki $\frac{d}{d x} sin x = cos x$ , hume milta hai $4 sin x$ . Yeh step kyun? Derivative ko ulta padho — cosine, sine ka derivative hai, toh sine, cosine ka antiderivative hai. Koi sign change nahi.
$- 3 sin x$ integrate karo. Hume $F^{'} = sin x$ chahiye. Hum jaante hain $\frac{d}{d x} cos x = - sin x$ , jo hume chahiye uska negative hai, toh $\int sin x d x = - cos x$ . Isliye $- 3 \cdot (- cos x) = + 3 cos x$ . Yeh step kyun? Minus cosine ke derivative par hai, toh yahan re-appear hota hai — aur aage $- 3$ ke saath combine hone par, do minuses milke ek plus ban jaate hain.
Combine karo. $4 sin x + 3 cos x + C .$

Verify karo: $\frac{d}{d x} (4 sin x + 3 cos x) = 4 cos x - 3 sin x$ ✓ — exactly wahi integrand. Extended table ke liye Trigonometric integrals dekho.

Neeche wala figure $sin x$ , $cos x$ , aur dashed $- cos x$ plot karta hai taaki tum dekh sako kahan minus sign paida hota hai: $cos x$ integrate karne par $sin x$ milta hai bina kisi flip ke, lekin $sin x$ integrate karne par dashed curve $- cos x$ milta hai, jo axis ke neeche exactly wahan baitha hai jahan tum $+ cos x$ expect karte.

Worked example Example 6 — Cell G: zero aur degenerate integrands

Do quick degenerate cases. (a) $\int 0 d x$ (b) $\int 7 d x .$

Forecast: agar tum "kuch nahi" integrate karo, toh answer kya ho sakta hai? Aur plain constant $7$ — isse $7 x^{0}$ treat karo.

Case (a): integrand $0$ hai. Hum poochte hain "kiska derivative $0$ hai?" Answer: koi bhi constant. Toh $\int 0 d x = C$ . Yeh step kyun? Differentiation har constant ko $0$ mein flatten kar deta hai; isse ulta chalane par, $0$ kisi bhi constant se aa sakta tha — yeh sabse pure illustration hai ki $+ C$ kyun exist karta hai.
Case (b): ek constant, power ki tarah likha. $7 = 7 x^{0}$ , toh $n = 0$ , $n + 1 = 1$ : $7 \cdot \frac{x ^{1}}{1} = 7 x$ . Yeh step kyun? Ek bare number secretly $x^{0}$ hota hai; power rule tab apply hota hai aur jaana-pehchaana fact confirm karta hai $\int k d x = k x + C$ .
Answers. (a) $C$ . (b) $7 x + C$ .

Verify karo: (a) $\frac{d}{d x} C = 0$ ✓. (b) $\frac{d}{d x} (7 x) = 7$ ✓.

Worked example Example 7 — Cell H: real-world word problem (velocity → position)

Pehle coordinate system set up karo. Maano $s (t)$ ball ki metres mein ground se upar height hai, upar ko positive maana gaya hai, $t$ seconds mein. Ball $s (0) = 20$ m height se rest par drop ki gayi hai aur gravity ke under neeche accelerate karti hai, toh uski velocity (upward-positive) hai $v (t) = - 9.8 t$ — negative kyunki ball neeche ja rahi hai. $s (t)$ nikalo.

Forecast: position, velocity ka antiderivative hai. $- 9.8 t$ integrate karne par $t^{2}$ term aayega ek leading number ke saath — aur ek $+ C$ jo hume starting height use karke pin down karna hoga.

Velocity integrate karke position nikalo. $s (t) = \int v (t) d t = \int - 9.8 t d t = - 9.8 \cdot \frac{t ^{2}}{2} + C = - 4.9 t^{2} + C .$ Yeh step kyun? Velocity, position ka derivative hai ( $v = d s / d t$ ), toh position, velocity ka antiderivative hai. $n = 1$ ke saath power rule se $\frac{t ^{2}}{2}$ milta hai, aur $- 9.8/2 = - 4.9$ . Wahi sign convention (upward positive) poore mein use ho raha hai, toh $- 4.9$ negative rehta hai.
Initial condition se $C$ pin down karo. $+ C$ hi starting height hai (yeh parent ke detective analogy ka "car ka unknown start" hai). $t = 0$ aur $s (0) = 20$ substitute karo: $s (0) = - 4.9 (0)^{2} + C = C ⟹ C = 20.$ Yeh step kyun? Hamare paas ek unknown ( $C$ ) aur ek jaana-maana fact ( $s (0) = 20$ ) hai; equation $C = 20$ solve karne se woh fix ho jaata hai. (Yahan $C$ height ke barabar hi nikla, $20$ , kyunki $t^{2}$ term $t = 0$ par vanish ho jaata hai.)
Position function likho. $s (t) = 20 - 4.9 t^{2} .$

Verify karo: $s (0) = 20 - 4.9 (0)^{2} = 20$ ✓ (units: metres). Differentiate karo: $\frac{d s}{d t} = \frac{d}{d t} (20 - 4.9 t^{2}) = - 9.8 t = v (t)$ ✓ (units: m/s, negative = downward, consistent). $t = 2$ par: $s = 20 - 4.9 (4) = 0.4$ m — almost land kar gaya. Limits of integration wala version ke liye Definite integrals & FTC dekho.

Worked example Example 8 — Cell I: exam twist, hidden power (pehle algebra)

$\int \frac{x ^{2} + 3 x - 1}{x} d x .$

Forecast: integration mein koi quotient rule nahi hota. Tumhe pehle divide karna hoga taaki har term ek plain power ban jaye. Teen exponents predict karo jo tumhe milenge.

Fraction ko $x = x^{1/2}$ par split karo. $\frac{x ^{2}}{x ^{1/2}} + \frac{3 x}{x ^{1/2}} - \frac{1}{x ^{1/2}} = x^{3/2} + 3 x^{1/2} - x^{- 1/2}$ . Yeh step kyun? Same bases divide karte waqt exponents subtract karo ( $2 - \frac{1}{2} = \frac{3}{2}$ , $1 - \frac{1}{2} = \frac{1}{2}$ , $0 - \frac{1}{2} = - \frac{1}{2}$ ). Yeh ek "product/quotient" ko clean sum of powers mein convert karta hai — woh ek hi form hai jo basic rules handle kar sakti hain. (Koi product/quotient rule exist nahi karta; uske liye Integration by parts ya Integration by substitution chahiye hogi.)
Har power integrate karo.
- $\int x^{3/2} d x = \frac{x ^{5/2}}{5/2} = \frac{2}{5} x^{5/2}$
- $\int 3 x^{1/2} d x = 3 \cdot \frac{x ^{3/2}}{3/2} = 2 x^{3/2}$
- $\int - x^{- 1/2} d x = - \frac{x ^{1/2}}{1/2} = - 2 x^{1/2}$ Yeh step kyun? Seedha power rule term by term; messy fractions $n + 1$ se divide karne se aate hain.
Combine karo. $\frac{2}{5} x^{5/2} + 2 x^{3/2} - 2 x^{1/2} + C .$

Verify karo: $\frac{d}{d x} (\frac{2}{5} x^{5/2} + 2 x^{3/2} - 2 x^{1/2}) = x^{3/2} + 3 x^{1/2} - x^{- 1/2}$ , jo recombine hokar $\frac{x ^{2} + 3 x - 1}{x}$ deta hai ✓.

Worked example Example 9 — Cell J: limiting edge, kyun

$n = - 1$ truly special hai Socho $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C$ aur $n$ ko $- 1$ ki taraf creep karne do. Kya break hota hai, aur kya take over karta hai?

Forecast: jaise-jaise $n \to - 1$ , denominator $n + 1 \to 0$ , toh formula blow up ho jaata hai — phir bhi actual integral $\int x^{- 1} d x$ perfectly finite hai ( $ln ∣ x ∣$ ). Yeh dono facts saath mein kaise rahte hain?

Denominator dekho. Jaise $n \to - 1$ , $n + 1 \to 0$ , toh $\frac{x ^{n + 1}}{n + 1}$ mein $\frac{1}{0}$ hai — power-rule expression genuinely diverge karta hai ek fixed nonzero $x \neq = 1$ ke liye. Yeh step kyun? Yeh dikhata hai ki $n \neq = - 1$ ka exclusion ek bolted-on technicality nahi hai — formula wahan literally evaluate hi nahi ho sakta.
Dekho gap kya fill karta hai. $n = - 1$ par correct antiderivative ek alag species ka function hai, $ln ∣ x ∣$ , $x$ ki koi power nahi. Yeh step kyun? Powers differentiate hone par kabhi $\frac{1}{x}$ nahi de sakti (unke derivatives phir powers hote hain), toh logarithm — ek bilkul naya function — required hai.
Conclusion. $\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + C (n \neq = - 1)$ , aur exactly excluded point par, $\int x^{- 1} d x = ln ∣ x ∣ + C$ .

Verify (blow-up ki numeric peek): $x = 2$ ke saath, $\frac{x ^{n + 1}}{n + 1}$ ko $n = - 0.9$ aur $n = - 0.99$ par evaluate karo: tumhe lagbhag $1.03$ phir $0.100$ times ten milta hai — more precisely $\approx 10.3$ aur $\approx 103$ — $n \to - 1$ jaane par unbounded badhta hua, divergence confirm karta hai.

Neeche wala figure us expression ko $n$ ke against plot karta hai: dekho purple curve coral dashed line $n = - 1$ ke paas aate hi rocket ki tarah upar jaata hai, jabki mint dashed line finite value $ln 2$ mark karti hai jo true integral excluded point par leta hai.

Recall Har cell ke liye ek-line reason

Positive power ::: raise & divide, koi sign fuss nahi. Negative power ::: raise & divide, lekin $n + 1$ negative ho sakta hai → minus sign aata hai. $n = - 1$ ::: power rule kaam nahi karta (÷0); $ln ∣ x ∣$ use karo. $\int sin x$ ::: $- cos x$ (minus cosine ke derivative par rehta hai). $\int a^{x}$ ::: $ln a$ se divide karo; base $e$ ko koi divisor nahi chahiye. Constant $k$ ::: $\int k d x = k x + C$ (yeh $k x^{0}$ hai). Word problem ::: $+ C$ initial condition hai — usse solve karo. Quotient/product integrand ::: pehle powers ka sum banao; koi quotient rule nahi.

Connections

Parent: Basic integration rules
Derivatives — basic rules (yahan har check answer differentiate karta hai)
Integration by substitution (jab integrand ek composite ho, plain power nahi)
Integration by parts (jab ek genuine product upar ke tricks se resist kare)
Definite integrals & FTC (Example 7 actual limits of integration ke saath)
Natural log and exponential functions ( $ln ∣ x ∣$ aur $\frac{a ^{x}}{l n a}$ rules ka source)
Trigonometric integrals (Example 4 ki table extend karta hai)