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Business Mathematics and Basic Statistics · Ch 14 — Integration

Standard Formula Integrals

2

Standard Formula Integrals

This WBCHSE Class 12 Commerce Business Mathematics and Basic Statistics chapter fixes a small set of standard-formula integrals as the toolkit every worked example draws from. Each one can be confirmed by differentiating the right-hand side and checking that the original integrand reappears.

Linearity of integration

Before listing the formulas, one property makes them usable together: integration is linear — the integral of a sum is the sum of the integrals, and a constant multiplier can be pulled outside the integral sign:

∫[a f(x)+b g(x)] dx=a∫f(x) dx+b∫g(x) dx\int \big[a\,f(x) + b\,g(x)\big]\,dx = a\int f(x)\,dx + b\int g(x)\,dx

This is exactly what lets a multi-term expression be integrated term by term, one standard formula at a time.

The five standard-formula integrals

#IntegralResultCondition
1∫xn dx\displaystyle\int x^{n}\,dxxn+1n+1+C\dfrac{x^{n+1}}{n+1}+Cn≠−1n \ne -1
2∫1x dx\displaystyle\int \frac{1}{x}\,dx$\lnx
3∫ex dx\displaystyle\int e^{x}\,dxex+Ce^{x}+C—
4∫1x2−a2 dx\displaystyle\int \frac{1}{x^{2}-a^{2}}\,dx$\dfrac{1}{2a}\ln\left\dfrac{x-a}{x+a}\right
5∫1a2−x2 dx\displaystyle\int \frac{1}{a^{2}-x^{2}}\,dx$\dfrac{1}{2a}\ln\left\dfrac{a+x}{a-x}\right
Note

Why formula 1 excludes n=−1n=-1

Substituting n=−1n=-1 into xn+1n+1\dfrac{x^{n+1}}{n+1} gives x00\dfrac{x^{0}}{0}, which is undefined (division by zero) — this is exactly the case formula 2 exists to cover separately, since ∫x−1 dx=∫1x dx=ln⁡∣x∣+C\displaystyle\int x^{-1}\,dx = \int \frac{1}{x}\,dx = \ln|x|+C.

Verifying formula 4 by differentiation

As a check on formula 4, differentiate the right-hand side using the chain rule and the log-quotient rule ln⁡∣u/v∣=ln⁡∣u∣−ln⁡∣v∣\ln|u/v| = \ln|u|-\ln|v|:

ddx[12a(ln⁡∣x−a∣−ln⁡∣x+a∣)]=12a(1x−a−1x+a)=12a⋅(x+a)−(x−a)(x−a)(x+a)=12a⋅2ax2−a2=1x2−a2\frac{d}{dx}\left[\frac{1}{2a}\Big(\ln|x-a|-\ln|x+a|\Big)\right] = \frac{1}{2a}\left(\frac{1}{x-a}-\frac{1}{x+a}\right) = \frac{1}{2a}\cdot\frac{(x+a)-(x-a)}{(x-a)(x+a)} = \frac{1}{2a}\cdot\frac{2a}{x^{2}-a^{2}} = \frac{1}{x^{2}-a^{2}}

which is exactly the original integrand — confirming the formula. Formula 5 checks the same way, with the sign of x2−a2x^{2}-a^{2} reversed throughout.

Reading the x>ax>a / a>xa>x conditions …

Definition 1Linearity of Integration

∫[a f(x)+b g(x)] dx=a ⁣∫f(x) dx+b ⁣∫g(x) dx\displaystyle\int\big[a\,f(x)+b\,g(x)\big]\,dx = a\!\int f(x)\,dx + b\!\int g(x)\,dx — a sum can be integrated term by term, and a constant factor can be …

Definition 2The Five Standard Integrals

∫xndx=xn+1n+1+C (n≠−1)\int x^{n}dx=\frac{x^{n+1}}{n+1}+C\ (n\ne-1); ∫1xdx=ln⁡∣x∣+C\int\frac{1}{x}dx=\ln|x|+C; ∫exdx=ex+C\int e^{x}dx=e^{x}+C; ∫1x2−a2dx=12aln⁡∣x−ax+a∣+C (x>a)\int\frac{1}{x^{2}-a^{2}}dx=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C\ (x>a); $\int\frac{1}{a^{2}-x^{2}}dx=\frac{1}{2a}\ …

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