### The basic integrals

Let us first recall the basic integrals we have learned (arbitrary constants are omitted):

- $\displaystyle \int x^n \; \mathrm{d}x = \frac{x^{n+1}}{n+1}$ for $n \neq -1$
- $\displaystyle \int \frac{1}{x} \; \mathrm{d}x = \ln |x|$
- $\displaystyle \int \mathrm{e}^x \; \mathrm{d}x = \mathrm{e}^x$
- $\displaystyle \int \sin x \; \mathrm{d}x = - \cos x$
- $\displaystyle \int \cos x \; \mathrm{d}x = \sin x$
- $\displaystyle \int \mathrm{sec}^2 x \; \mathrm{d}x = \tan x$
- $\displaystyle \int a f(x) \pm b g(x) \; \mathrm{d}x = a \int f(x) \; \mathrm{d}x \pm b \int g(x) \; \mathrm{d}x$

where $a$ and $b$ are constants

They should hopefully be familiar to us, with perhaps the exception of the modulus in the $\ln x$ result.

The reason for the modulus is because $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} ( \ln x ) = \frac{1}{x}$ and, by the chain rule, $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \big ( \ln (- x) \big ) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$. We use the first expression when $x$ is positive and expressions like $\ln (-x)$ may arise when $x$ is negative. Hence, depending on the domain, integration of $\frac{1}{x}$ gives us either $\ln x$ or $\ln (-x)$. These two cases can be combined into $\displaystyle \int \frac{1}{x} \; \mathrm{d}x = \ln |x| + C$.

As a general rule, always include the modulus unless we are certain the expression within the logarithm is positive.

### Linear modifications

All the formulas above can be modified slightly. Instead of $\displaystyle \int x^5 \; \mathrm{d}x = \frac{x^6}{6} + C$, we may be asked to find $\displaystyle \int (2x-3)^5 \; \mathrm{d}x$.

Recall the chain rule for differentiation of expressions such as $f(x) = (2x-3)^6$. In step one, we differentiate as per usual to get $6(2x-3)^5$. But we are not done yet. We must differentiate "inside" to get 2, leaving us with the final answer $f'(x)=6 (2x-3)^5 \cdot 2 = 12 (2x-3)^5$. In other words, we have to further multiply by the coefficient of $x$ at the final step.

For integration, we reverse the process and divide by the coefficient of $x$ at the final step. This explains why $\displaystyle \int \frac{1}{2x-3} \; \mathrm{d}x = \frac{1}{2} | 2x - 3 | + C$ and $\displaystyle \int \cos (3-2x) \; \mathrm{d}x = \frac{\sin (3-2x)}{-2} + C$ .

A word of caution: reversing chain rule and dividing by the coefficient of $x$ only works for linear expressions of the form $mx+c$. Anything more complicated cannot be done by an analgous procedure. For example, $\int (x^2 + 3)^{\frac{1}{2}} \; \mathrm{d}x$ IS NOT $\displaystyle \frac{(x^2+3)^{\frac{3}{2}} }{\frac{3}{2} (2x) }$. This rule only work for numbers, so be wary if we find ourselves dividing by variable terms such as $x, x^2$ or even more complicated expressions. It is likely we have made a mistake! (If you are interested, discuss with your teacher/a friend/me why this process works for numbers but not $x$'s.)

So how do we find $\int (x^2 + 3)^{\frac{1}{2}} \; \mathrm{d}x$. It turns out we won't be required to do this without hints for the "A" level syllabus. Building off the idea of reversing chain rule for differentiation, however, we will be required to know how to find $\int x (x^2 + 3)^{\frac{1}{2}} \; \mathrm{d}x$. We will discuss the technique for doing this in example 8.1.