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

Kaunsa rule? Power rule: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$. Kyun fire karta hai: integrand ek plain power of $x$ hai jisme $n=5$ hai, aur $5\ne -1$, isliye hum $n+1$ se divide karne ki permission rakhte hain. Power ko ek badhao ($5\to 6$) aur us naye power se divide karo: $\int x^{5}dx=\frac{x^6}{6}+C.$ Verify: $\frac{d}{dx}\frac{x^6}{6}=\frac{6x^5}{6}=x^5.$ ✓

Kaunsa rule? Exponential rule. $e^x$ differentiation ka fixed point hai ($\frac{d}{dx}{e}^x=e^x$), isliye ulta padhne par bhi woh unchanged rehta hai. $\int e^{x}dx=e^x+C.$ Verify: $\frac{d}{dx}(e^x+C)=e^x.$ ✓

Kaunsa rule? Trig. Hum chahte hain $F$ aisa ho ki $F^{\prime }=\cos x$. Kyunki $\frac{d}{dx}\sin x=\cos x$, toh $F=\sin x$ lo. $\int \cos xdx=\sin x+C.$ Verify: $\frac{d}{dx}\sin x=\cos x.$ ✓

Kyun rewrite karein? Power rule ko $x$ ki actual power chahiye, fraction nahi. Rewrite karo $\frac{1}{x^4}=x^{-4}$. Ab $n=-4$ hai (aur $-4\ne -1$, isliye allowed hai). Badhao: $-4\to -3$; $-3$ se divide karo: $\int x^{-4}dx=\frac{x^{-3}}{-3}+C=-\frac{1}{3x^3}+C.$ Verify: $\frac{d}{dx}\left(-\frac{1}{3}{x}^{-3}\right)=-\frac{1}{3}(-3)x^{-4}=x^{-4}.$ ✓

Kyun rewrite karein? Kisi square ka cube-root ek chhupa hua power hai: $\sqrt[3]{x^2}=x^{2/3}$. Yahan $n=\frac{2}{3}$ hai, isliye $n+1=\frac{5}{3}$. $\frac{5}{3}$ se divide karo (= $\frac{3}{5}$ se multiply karo): $\int x^{2/3}dx=\frac{x^{5/3}}{5/3}+C=\frac{3}{5}{x}^{5/3}+C.$ Verify: $\frac{d}{dx}\frac{3}{5}{x}^{5/3}=\frac{3}{5}\cdot \frac{5}{3}{x}^{2/3}=x^{2/3}.$ ✓

Kaunsa rule? General-base exponential. Natural log and exponential functions se hum jaante hain $\frac{d}{dx}{5}^x=5^x\ln 5$ — yeh ulta padhne par $\ln 5$ times zyada ho jaata hai. $\ln 5$ se kyun divide karein? Woh extra factor cancel karne ke liye jo derivative introduce karta hai: $\int 5^{x}dx=\frac{5^x}{\ln 5}+C.$ Verify: $\frac{d}{dx}\frac{5^x}{\ln 5}=\frac{5^x\ln 5}{\ln 5}=5^x.$ ✓

Power rule kyun nahi? $\frac{1}{x}=x^{-1}$ likhne par $n=-1$ milta hai, aur power rule $n+1=0$ se divide karta hai — jo undefined hai. Yahi woh hole hai jiske baare mein parent note ne warning di thi. Us hole ko kya fill karta hai? Woh function jiska derivative $\frac{1}{x}$ hai woh hai $\ln |x|$. Absolute value important hai: $\frac{1}{x}$ negative $x$ ke liye bhi exist karta hai, aur $\frac{d}{dx}\ln |x|=\frac{1}{x}$ sabhi $x\ne 0$ ke liye. Pehle constant bahar nikalo (linearity): $\int \frac{7}{x}dx=7\int \frac{1}{x}dx=7\ln |x|+C.$ Verify: $\frac{d}{dx}7\ln |x|=7\cdot \frac{1}{x}=\frac{7}{x}.$ ✓

Sign kahan rehta hai: term by term integrate karo (linearity). $\int \sin xdx=-\cos x$ — minus $\frac{d}{dx}\cos x=-\sin x$ se aata hai, jise hum flip karte hain. $\int \cos xdx=\sin x$ — yahan koi minus nahi. Toh $\int (\sin x-\cos x)dx=-\cos x-\sin x+C.$ Verify: $\frac{d}{dx}(-\cos x-\sin x)=\sin x-\cos x.$ ✓

Pehle expand kyun karein? Integration ke liye koi product rule nahi hai, isliye hum bracket ko product ki tarah integrate nahi kar sakte. Ise ek sum mein multiply karke kholein jise table recognize kare: $\sec x(\sec x+\tan x)={\sec }^{2}x+\sec x\tan x.$ Ab har piece ek jaana-pehchana reverse-derivative hai: $\int {\sec }^{2}xdx=\tan x$ (kyunki $\frac{d}{dx}\tan x={\sec }^{2}x$), $\int \sec x\tan xdx=\sec x$ (kyunki $\frac{d}{dx}\sec x=\sec x\tan x$). $\int \sec x(\sec x+\tan x)dx=\tan x+\sec x+C.$ Verify: $\frac{d}{dx}(\tan x+\sec x)={\sec }^{2}x+\sec x\tan x=\sec x(\sec x+\tan x).$ ✓

Strategy: linearity hume sum split karne aur har constant bahar kheenchne deti hai; phir har term ek single rule hai.

• $\int 4x^{3}dx=4\cdot \frac{x^4}{4}=x^4$ (power rule).
• $\int -\frac{2}{x}dx=-2\ln |x|$ (log rule — $n=-1$).
• $\int 3e^{x}dx=3e^x$ (exponential, unchanged).
• $\int -5\sin xdx=+5\cos x$ ($\int \sin =-\cos$ se sign flip).

$\int \left(4x^3-\frac{2}{x}+3e^x-5\sin x\right)dx=x^4-2\ln |x|+3e^x+5\cos x+C.$ Verify: differentiate karo: $4x^3-\frac{2}{x}+3e^x-5\sin x.$ ✓

Definite integral ka matlab (Definite integrals & FTC se): pehle koi bhi antiderivative $F$ dhundho, phir $F(\text{top})-F(\text{bottom})$ calculate karo. $+C$ cancel ho jaata hai kyunki woh dono mein appear karta hai aur subtract ho jaata hai — isliye hum definite integrals mein use carry nahi karte. Power rule se antiderivative: $F(x)=\frac{x^3}{3}$. Limits ke beech evaluate karo ($[{]}_1^2$ notation ka matlab "top minus bottom"): ${\int }_1^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}.$ Yeh kaisa dikhta hai: figure mein $x=1$ se $x=2$ tak $y=x^2$ ke neeche ka red region exactly $\frac{7}{3}\approx 2.33$ area ka hai — yeh number width $1$ se bada hai kyunki curve us strip mein $y=1$ se kaafi upar baith ti hai. Verify: $\frac{d}{dx}\frac{x^3}{3}=x^2$, isliye $F$ ek valid antiderivative hai. ✓

Fraction split kyun karein? Integration ke liye koi quotient rule nahi hai. Lekin yeh quotient har top term ko $x$ se divide karke ek sum mein toot jaata hai: $\frac{x^2+3x-1}{x}=x+3-\frac{1}{x}.$ Ab term by term integrate karo:

• $\int xdx=\frac{x^2}{2}$,
• $\int 3dx=3x$,
• $\int -\frac{1}{x}dx=-\ln |x|$ ($n=-1$ patch).

$\int \frac{x^2+3x-1}{x}dx=\frac{x^2}{2}+3x-\ln |x|+C.$ Verify: $\frac{d}{dx}\left(\frac{x^2}{2}+3x-\ln |x|\right)=x+3-\frac{1}{x}=\frac{x^2+3x-1}{x}.$ ✓

Teen alag rules, ek integral. Split karo (linearity), phir:

• $\int 2^{x}dx=\frac{2^x}{\ln 2}$ (general-base exponential — $\ln 2$ divide out karo).
• $\int {\sec }^{2}xdx=\tan x$ ($\frac{d}{dx}\tan x={\sec }^{2}x$ ka reverse).
• $\frac{4}{\sqrt{x}}=4x^{-1/2}$, isliye $\int 4x^{-1/2}dx=4\cdot \frac{x^{1/2}}{1/2}=8x^{1/2}=8\sqrt{x}.$

$\int \left(2^x+{\sec }^{2}x+\frac{4}{\sqrt{x}}\right)dx=\frac{2^x}{\ln 2}+\tan x+8\sqrt{x}+C.$ Verify: $\frac{d}{dx}\frac{2^x}{\ln 2}=2^x$; $\frac{d}{dx}\tan x={\sec }^{2}x$; $\frac{d}{dx}8x^{1/2}=8\cdot \frac{1}{2}{x}^{-1/2}=4x^{-1/2}=\frac{4}{\sqrt{x}}.$ ✓

Confusion kya hai: unhone neeche $x$ wala fraction dekha aur log rule pakad liya. Check expose karta hai: log rule sirf exponent exactly $-1$ ke liye hai. Yahan $\frac{1}{x^2}=x^{-2}$, aur $-2\ne -1$, isliye ordinary power rule apply hota hai: $\int x^{-2}dx=\frac{x^{-1}}{-1}+C=-\frac{1}{x}+C.$ Correct answer verify karo: $\frac{d}{dx}\left(-x^{-1}\right)=x^{-2}=\frac{1}{x^2}.$ ✓ Unka galat answer fail karta hai verify mein: $\frac{d}{dx}\ln |x^2|=\frac{1}{x^2}\cdot 2x=\frac{2}{x}\ne \frac{1}{x^2}.$ Mismatch hi tell hai — answer differentiate karne par integrand wapas aana chahiye, aur woh nahi aaya.