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

WHAT rule? The power rule: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$. WHY it fires: the integrand is a plain power of $x$ with $n=5$, and $5\ne -1$, so we are allowed to divide by $n+1$. Raise the power by one ($5\to 6$) and divide by that new power: $\int x^{5}dx=\frac{x^6}{6}+C.$ Verify: $\frac{d}{dx}\frac{x^6}{6}=\frac{6x^5}{6}=x^5.$ ✓

WHAT rule? The exponential rule. $e^x$ is the fixed point of differentiation ($\frac{d}{dx}{e}^x=e^x$), so read backwards it is unchanged too. $\int e^{x}dx=e^x+C.$ Verify: $\frac{d}{dx}(e^x+C)=e^x.$ ✓

WHAT rule? Trig. We want $F$ with $F^{\prime }=\cos x$. Since $\frac{d}{dx}\sin x=\cos x$, take $F=\sin x$. $\int \cos xdx=\sin x+C.$ Verify: $\frac{d}{dx}\sin x=\cos x.$ ✓

WHY rewrite? The power rule needs an actual power of $x$, not a fraction. Rewrite $\frac{1}{x^4}=x^{-4}$. Now $n=-4$ (and $-4\ne -1$, so we are allowed). Raise: $-4\to -3$; divide by $-3$: $\int x^{-4}dx=\frac{x^{-3}}{-3}+C=-\frac{1}{3x^3}+C.$ Verify: $\frac{d}{dx}\left(-\frac{1}{3}{x}^{-3}\right)=-\frac{1}{3}(-3)x^{-4}=x^{-4}.$ ✓

WHY rewrite? A cube-root of a square is a power in disguise: $\sqrt[3]{x^2}=x^{2/3}$. Here $n=\frac{2}{3}$, so $n+1=\frac{5}{3}$. Divide by $\frac{5}{3}$ (= multiply by $\frac{3}{5}$): $\int x^{2/3}dx=\frac{x^{5/3}}{5/3}+C=\frac{3}{5}{x}^{5/3}+C.$ Verify: $\frac{d}{dx}\frac{3}{5}{x}^{5/3}=\frac{3}{5}\cdot \frac{5}{3}{x}^{2/3}=x^{2/3}.$ ✓

WHAT rule? General-base exponential. From Natural log and exponential functions we know $\frac{d}{dx}{5}^x=5^x\ln 5$ — that is $\ln 5$ times too big when read backwards. WHY divide by $\ln 5$? To cancel that extra factor the derivative introduces: $\int 5^{x}dx=\frac{5^x}{\ln 5}+C.$ Verify: $\frac{d}{dx}\frac{5^x}{\ln 5}=\frac{5^x\ln 5}{\ln 5}=5^x.$ ✓

WHY not the power rule? Writing $\frac{1}{x}=x^{-1}$ gives $n=-1$, and the power rule divides by $n+1=0$ — undefined. This is exactly the hole the parent note warned about. What fills the hole? The function whose derivative is $\frac{1}{x}$ is $\ln |x|$. The absolute value matters: $\frac{1}{x}$ exists for negative $x$ too, and $\frac{d}{dx}\ln |x|=\frac{1}{x}$ for all $x\ne 0$. Pull the constant out first (linearity): $\int \frac{7}{x}dx=7\int \frac{1}{x}dx=7\ln |x|+C.$ Verify: $\frac{d}{dx}7\ln |x|=7\cdot \frac{1}{x}=\frac{7}{x}.$ ✓

WHERE the sign lives: integrate term by term (linearity). $\int \sin xdx=-\cos x$ — the minus comes from $\frac{d}{dx}\cos x=-\sin x$, which we flip. $\int \cos xdx=\sin x$ — no minus here. So $\int (\sin x-\cos x)dx=-\cos x-\sin x+C.$ Verify: $\frac{d}{dx}(-\cos x-\sin x)=\sin x-\cos x.$ ✓

WHY expand first? There is no product rule for integration, so we cannot integrate the bracket as a product. Multiply it out into a sum the table recognises: $\sec x(\sec x+\tan x)={\sec }^{2}x+\sec x\tan x.$ Now each piece is a known reverse-derivative: $\int {\sec }^{2}xdx=\tan x$ (since $\frac{d}{dx}\tan x={\sec }^{2}x$), $\int \sec x\tan xdx=\sec x$ (since $\frac{d}{dx}\sec x=\sec x\tan x$). $\int \sec x(\sec x+\tan x)dx=\tan x+\sec x+C.$ Verify: $\frac{d}{dx}(\tan x+\sec x)={\sec }^{2}x+\sec x\tan x=\sec x(\sec x+\tan x).$ ✓

Strategy: linearity lets us split the sum and pull each constant out; then each term is a single rule.

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

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

WHAT a definite integral means (from Definite integrals & FTC): first find any antiderivative $F$, then compute $F(\text{top})-F(\text{bottom})$. The $+C$ cancels because it appears in both and subtracts away — that is why we don't carry it in definite integrals. Antiderivative by the power rule: $F(x)=\frac{x^3}{3}$. Evaluate between the limits (the notation $[{]}_1^2$ means "top minus bottom"): ${\int }_1^2{x}^{2}dx={\left[\frac{x^3}{3}\right]}_1^2=\frac{8}{3}-\frac{1}{3}=\frac{7}{3}.$ What it looks like: the red region under $y=x^2$ from $x=1$ to $x=2$ in the figure has area exactly $\frac{7}{3}\approx 2.33$ — a number bigger than the width $1$ because the curve sits well above $y=1$ across that strip. Verify: $\frac{d}{dx}\frac{x^3}{3}=x^2$, so $F$ is a valid antiderivative. ✓

WHY split the fraction? There is no quotient rule for integration. But this quotient breaks apart into a sum by dividing each top term by $x$: $\frac{x^2+3x-1}{x}=x+3-\frac{1}{x}.$ Now integrate term by term:

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

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

Three different rules, one integral. Split (linearity), then:

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

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

The confusion: they saw a fraction with $x$ on the bottom and reached for the log rule. The check exposes it: the log rule is only for exponent exactly $-1$. Here $\frac{1}{x^2}=x^{-2}$, and $-2\ne -1$, so the ordinary power rule applies: $\int x^{-2}dx=\frac{x^{-1}}{-1}+C=-\frac{1}{x}+C.$ Verify the correct answer: $\frac{d}{dx}\left(-x^{-1}\right)=x^{-2}=\frac{1}{x^2}.$ ✓ Verify their wrong answer fails: $\frac{d}{dx}\ln |x^2|=\frac{1}{x^2}\cdot 2x=\frac{2}{x}\ne \frac{1}{x^2}.$ The mismatch is the tell — differentiating the answer must return the integrand, and it didn't.