Techniques of integration

Why the power −1 needs a logarithm

Look for the derivative of the denominator in the numerator. If the match differs by a constant factor, compensate outside the integral. If it differs by a function of $x$, the standard logarithm rule cannot be applied directly.

What happens if $n=-1$ in the power rule? Substitution would give $x^0/0+C$, which is undefined; it is a demonstration that the rule cannot apply, not an antiderivative.

For $f(x)>0$, $\frac{d}{dx}\ln|f(x)|=\frac{d}{dx}\ln f(x)=f^{\prime}(x)/f(x)$. For $f(x)<0$, differentiate $\ln[-f(x)]$: the two minus signs give $[-f^{\prime}(x)]/[-f(x)]=f^{\prime}(x)/f(x)$. Thus on intervals where $f\ne0$,

$$\int\frac{f^{\prime}(x)}{f(x)}\,dx=\ln|f(x)|+C.$$

The basic case is $\int1/x\,dx=\ln|x|+C$. The Teacher labels the basic rule O-level and its composite form H2-level. Inspect the inner derivative first; supply only a constant multiple with compensation. For an improper rational function, use long division before choosing a standard form. Examples 2 and 3 show these choices.

Exponential standard forms

Reverse $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}e^{f(x)}=f^{\prime}(x)e^{f(x)}$: the basic integral is $\int e^x\,dx=e^x+C$ and the composite integral is $\int f^{\prime}(x)e^{f(x)}\,dx=e^{f(x)}+C$. The Teacher labels the basic form O-level and the composite form H2-level.

For a constant base, $\frac{d}{dx}a^x=a^x\ln a$ and $\frac{d}{dx}a^{f(x)}=f^{\prime}(x)a^{f(x)}\ln a$. Hence $\int a^x\,dx=a^x/\ln a+C$ and $\int f^{\prime}(x)a^{f(x)}\,dx=a^{f(x)}/\ln a+C$, for $a>0$, $a\ne1$. The printed table states $a>0$ but omits $a\ne1$; division by $\ln1=0$ is undefined. When $a=1$, integrate $1$ to $x+C$, or $f^{\prime}(x)$ to $f(x)+C$. This qualification supplements the preserved source table. Example 4(e) uses the base 5 rule.

Inverse-trigonometric standard forms

Here $\sin^{-1}$ and $\tan^{-1}$ mean inverse functions. Reverse $\frac{d}{dx}\sin^{-1}x=1/\sqrt{1-x^2}$ and $\frac{d}{dx}\tan^{-1}x=1/(1+x^2)$ to obtain their integrals, each with $C$. For $a>0$:

Basic form: $\int\frac1{x^2+a^2}\,dx=\frac1a\tan^{-1}(x/a)+C$

Composite form: $\int\frac{f^{\prime}(x)}{[f(x)]^2+a^2}\,dx=\frac1a\tan^{-1}[f(x)/a]+C$

Basic form: $\int\frac1{\sqrt{a^2-x^2}}\,dx=\sin^{-1}(x/a)+C$, $\lvert x\rvert<a$

Composite form: $\int\frac{f^{\prime}(x)}{\sqrt{a^2-[f(x)]^2}}\,dx=\sin^{-1}[f(x)/a]+C$, $\lvert f(x)\rvert<a$

The source marks the basic formulas as available in MF27. The Teacher emphasises numerator 1 and squared-variable coefficient 1 for a basic form; a composite form matches its inner derivative.

The source says conditions need not be written in the working, while the expression must still be well-defined.

In the radical denominator, $a^2-[f(x)]^2>0$ gives $-a<f(x)<a$, equivalently $\lvert f(x)\rvert<a$. Endpoints where that denominator is zero are excluded.

Example 5(c) preserves both the basic-form and composite-form methods.

Complete the square to expose a standard form

For a non-factorisable reciprocal quadratic $1/(px^2+qx+r)$ or a reciprocal radical $1/\sqrt{-px^2+qx+r}$, with $p>0$, the printed remark instructs completing the square.

First factor the squared-variable coefficient, then compare coefficients in $(x+a)^2+b$ if necessary.

Example 6 uses shifts with derivative 1; a factor under a radical emerges as its square root, rather than as itself.

Preserve a positive radicand on the chosen interval. This remark is a method for the displayed forms, not a claim that every quadratic has the same arctangent antiderivative.

Rational logarithmic forms and partial fractions

1. Compare the numerator and denominator degrees; divide first if needed.
2. Factor the denominator and include every repeated-factor power.
3. Solve for the partial-fraction coefficients, then integrate each term.
4. Keep the denominator restrictions and differentiate the final answer to check it.

For $a>0$, the source gives $\int1/(x^2-a^2)\,dx=\frac1{2a}\ln[(x-a)/(x+a)]+C$ on $x>a$, and its composite version $\int f^{\prime}(x)/([f(x)]^2-a^2)\,dx=\frac1{2a}\ln[(f(x)-a)/(f(x)+a)]+C$ on $f(x)>a$. It also gives $\int1/(a^2-x^2)\,dx=\frac1{2a}\ln[(a+x)/(a-x)]+C$ on $|x|<a$, and the composite version $\int f^{\prime}(x)/(a^2-[f(x)]^2)\,dx=\frac1{2a}\ln[(a+f(x))/(a-f(x))]+C$ on $|f(x)|<a$. The basic formulas are marked MF27. The Teacher highlights using moduli inside logarithms when a sign restriction is not specified. With moduli these formulas hold on any connected interval avoiding the denominator zeros; moduli do not remove poles.

The source recalls three partial-fraction patterns (with distinct factors and nonzero denominators):

• $\frac{px+q}{(ax+b)(cx+d)}=\frac A{ax+b}+\frac B{cx+d}$.
• $\frac{px^2+qx+r}{(ax+b)(cx+d)^2}=\frac A{ax+b}+\frac B{cx+d}+\frac C{(cx+d)^2}$: retain every power of a repeated linear factor.
• $\frac{px^2+qx+r}{(ax+b)(x^2+c^2)}=\frac A{ax+b}+\frac{Bx+C}{x^2+c^2}$: an irreducible quadratic needs a linear numerator.

The Teacher emphasises starting with a proper rational fraction and recognising factor structure; divide first if it is improper. The source labels these templates MF27.

To derive the first logarithmic formula, write $1/(x^2-a^2)=P/(x-a)+Q/(x+a)$. Clearing denominators gives $1=P(x+a)+Q(x-a)$. Substituting $x=a$ gives $P=1/(2a)$ and $x=-a$ gives $Q=-1/(2a)$.

Thus integration gives $[\ln|x-a|-\ln|x+a|]/(2a)+C=\frac1{2a}\ln\left|\frac{x-a}{x+a}\right|+C$.

The displayed logarithm law is $\ln a-\ln b=\ln(a/b)$ for positive arguments; applying it to absolute values preserves the needed signs. Self-Practice 6 asks for the companion proof.

If a numerator is not directly proportional to the denominator derivative, split it into that derivative plus a remainder. Example 9 uses $6x+16=3(2x+4)+4$. Example 10 preserves two methods and the Teacher note explaining why their different-looking logarithms agree.

Trigonometric standard forms and identities

Check the inner derivative before choosing a formula. If no direct match appears, simplify with an identity: lower even sine/cosine powers, or replace tangent/cotangent squares using the corresponding secant/cosecant identity. Recheck signs and chain-rule factors after integrating.

Reverse $\frac{d}{dx}\sin[f(x)]=f^{\prime}(x)\cos[f(x)]$ to get the first composite rule below. The same reasoning gives all six source pairs (each with an arbitrary $C$):

Basic integral: $\int\cos x\,dx=\sin x+C$

Composite integral: $\int f^{\prime}(x)\cos[f(x)]\,dx=\sin[f(x)]+C$

Basic integral: $\int\sin x\,dx=-\cos x+C$

Composite integral: $\int f^{\prime}(x)\sin[f(x)]\,dx=-\cos[f(x)]+C$

Basic integral: $\int\sec^2x\,dx=\tan x+C$

Composite integral: $\int f^{\prime}(x)\sec^2[f(x)]\,dx=\tan[f(x)]+C$

Basic integral: $\int\csc^2x\,dx=-\cot x+C$

Composite integral: $\int f^{\prime}(x)\csc^2[f(x)]\,dx=-\cot[f(x)]+C$

Basic integral: $\int\sec x\tan x\,dx=\sec x+C$

Composite integral: $\int f^{\prime}(x)\sec[f(x)]\tan[f(x)]\,dx=\sec[f(x)]+C$

Basic integral: $\int\csc x\cot x\,dx=-\csc x+C$

Composite integral: $\int f^{\prime}(x)\csc[f(x)]\cot[f(x)]\,dx=-\csc[f(x)]+C$

The source writes cosec for csc. The Teacher labels the basic forms O-level and the composite forms H2-level, emphasising the inner derivative and the minus signs in the sine/cosecant cases; the six composite rules are annotated as not in MF27.

Four logarithmic formulas are marked available in MF27: $\int\tan x\,dx=\ln(\sec x)+C$ and $\int\sec x\,dx=\ln(\sec x+\tan x)+C$ on $|x|<\pi/2$; $\int\cot x\,dx=\ln(\sin x)+C$ and $\int\csc x\,dx=-\ln(\csc x+\cot x)+C$ on $0<x<\pi$. The source remark extends these using absolute values inside the logarithms; work on connected intervals where the original trigonometric functions are defined.

For even powers of sine/cosine, lower the power with $\sin2A=2\sin A\cos A$ and $\cos2A=2\cos^2A-1=1-2\sin^2A$, marked MF27. Rearranging gives $\cos^2A=(1+\cos2A)/2$ and $\sin^2A=(1-\cos2A)/2$. Examples 12–13 use these, applying double angle twice for the fourth power. The Teacher crosses out the false power-rule answer $\cos^3x/3+C$ to $\int\cos^2x\,dx$: its derivative is $-\cos^2x\sin x$, not $\cos^2x$.

Other printed identities are $\cos^2A+\sin^2A=1$, $1+\tan^2A=\sec^2A$ and $1+\cot^2A=\csc^2A$. The Teacher adds a worked demonstration: $\int\tan^2x\,dx=\int(\sin^2x/\cos^2x)\,dx=\int(\sec^2x-1)\,dx=\tan x-x+C$, with $\frac{d}{dx}\tan x=\sec^2x$. Keep $\cos x\ne0$ on its interval. This is a Teacher annotation, not a new printed Student exercise. Example 15 retains both supplied product-to-sum identities.

Why substitution works

Use one variable throughout the transformed integral:

1. Differentiate the stated substitution to obtain the new differential.
2. Replace the integrand and differential completely, keeping any sign or interval restriction.
3. Integrate in the new variable.
4. Return to the original variable and differentiate to check.

A square root is nonnegative. A trigonometric substitution must use an angle interval that justifies the sign when simplifying it.

The source contrasts $\int1/\sqrt{1-4x^2}\,dx$, which matches the composite arcsine formula after supplying the inner derivative, with the motivating question $\int\sqrt{1-x^2}\,dx$, which does not match that reciprocal-radical form directly. A suitable change of variable can turn it into standard integrals.

Let $x=g(u)$, $dx/du=g^{\prime}(u)$ and $y=f(x)$. If $\int y\,dx=\int f(x)\,dx=F(x)+C$, then $F^{\prime}(x)=f(x)$. The chain rule gives $\frac{d}{du}F(g(u))=F^{\prime}(g(u))g^{\prime}(u)=f(g(u))g^{\prime}(u)$. Integrating with respect to $u$ therefore gives

$$\int f(x)\,dx=\int f(g(u))g^{\prime}(u)\,du.$$

An additive constant is absorbed in the resulting indefinite integral. Every occurrence of $x$ in the transformed integrand must be rewritten in terms of $u$, including the differential. The Teacher highlights this change of variable and corrects an incomplete handwritten $\int f(g(\theta))\,d\theta$ by inserting $dx/d\theta$; using $x=g(\theta)$ requires $dx=(dx/d\theta)\,d\theta$. The margin chain-rule reminder is $d(y^2)/dx=2y\,dy/dx$. Apply this derivation in Example 16 and Self-Practice 9 below.

Parts reverses the product rule

Choose the differentiated factor so that the remaining integral becomes simpler. Compute its derivative and the other factor’s antiderivative separately, then apply parts. Inspect the new integral before repeating: some problems return to the original integral, which can then be solved algebraically.

The product rule is $d(uv)/dx=u\,dv/dx+v\,du/dx$. Integrate and rearrange to obtain $\int u\,dv/dx\,dx=uv-\int v\,du/dx\,dx$. Choose $u$ to differentiate and $dv/dx$ to integrate. The source highlights inverse-trigonometric functions, logarithms and products of mixed functions. A standalone function can be multiplied by 1.

Compare the source two choices for $\int xe^x\,dx$. With $u=x$, $dv/dx=e^x$, $du/dx=1$, $v=e^x$, parts gives $xe^x-\int e^x\,dx=xe^x-e^x+C$. With $u=e^x$, $dv/dx=x$, $du/dx=e^x$, $v=x^2/2$, it gives $x^2e^x/2-\frac12\int x^2e^x\,dx$, which is harder. Choose the route that simplifies the remaining integral, sometimes by trial and error.

The decreasing preference LIATE is Logarithmic, Inverse trigonometric, Algebraic (including constants), Trigonometric, Exponential. The Teacher warns this is a guide with exceptions, not an absolute rule. Margin examples include $\ln(2x-1)$ and $\log_5x$, $\tan^{-1}x$ and $\sin^{-1}(2x)$, a polynomial and $4=4x^0$, trigonometric functions, $e^x$ and $2^x$. The Teacher crosses out confusing an inverse-sine integral with its derivative; multiplying the integrand by 1 makes parts available, as in Example 17(c).

The printed remarks refer to Example 16(c), Example 16(d,e) and then Example 17 after Example 18. In the current numbering these demonstrations are Example 17(c), Example 17(d,e), and Example 18 respectively. Their teaching points are preserved: introduce 1, repeat parts, and use a preceding derivative. Native numbering is retained in the source transcription.

A method checklist from the closing Teacher notes

For substitution, follow the four steps in Example 16(a): differentiate the relation; replace the integrand and differential completely; integrate in the new variable; return to the original variable. In Example 16(b), implicit differentiation of $u^2=x-2$ makes $dx/du=2u$ clear; use long division after substitution. A triangle recovers trigonometric ratios, with signed identities covering negative inputs. The two square-root substitutions in Example 16 use the stated angle intervals to control the radical sign.

The Teacher mastery notes add long division for improper fractions and rewriting a denominator as a power: $\int x/\sqrt{x^2-1}\,dx=\frac12\int2x(x^2-1)^{-1/2}\,dx=\sqrt{x^2-1}+C$ on $|x|>1$. The final route-map annotations reinforce that $\int\ln x\,dx\ne1/x+C$; instead parts gives $x\ln x-x+C$ for $x>0$. They contrast the reverse-chain integral $\frac12\int2xe^{x^2}\,dx=\frac12e^{x^2}+C$ with the product $\int xe^x\,dx$, which uses parts as above. Self-Practice 9 answers the earlier motivating radical integral in both sine and cosine substitutions.