Recognise standard forms

Choose a method from the structure: a derivative beside a composite suggests substitution; a polynomial multiplied by an exponential or logarithm often suggests parts. Differentiate your result to check it.

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.