The fundamental *use* of *integration* is as a *continuous
version of summing*. But, paradoxically, often integrals are *computed* by viewing integration as essentially an *inverse operation to differentiation*. (That fact is the so-called *Fundamental Theorem of Calculus*.)

The notation, which we're stuck with for historical reasons,
is as peculiar as the notation for derivatives: the **integral of a
function $f(x)$ with respect to $x$** is written as
$$\int f(x)\;dx$$

The remark that integration is (almost) an inverse to the operation of
differentiation means that if
$${d\over dx}f(x)=g(x)$$
then
$$\int g(x)\;dx=f(x)+C$$
The extra $C$, called the **constant of integration**, is really
necessary, since after all differentiation kills off constants, which
is why integration and differentiation are not *exactly* inverse
operations of each other.

Since integration is *almost* the inverse operation of
differentiation, recollection of formulas and processes for *differentiation* already tells the most important formulas for *integration:*
\begin{align*}\int x^n\; dx &= {1\over n+1}x^{n+1}+C &
\hbox{ unless $n=-1$ }\\
\int e^x \;dx&= e^x+C \\
\int {1\over x} \;dx&= \ln x+C \\
\int \sin x\;dx&=-\cos x+C \\
\int \cos x\;dx&= \sin x + C\\
\int \sec^2 x\;dx&=\tan x+C \\
\int {1\over 1+x^2} \; dx&=\arctan x+C
\end{align*}

And since the derivative of a sum is the sum of the derivatives, the
*integral of a sum is the sum of the integrals:*
$$ \int f(x)+g(x)\;dx=\int f(x)\;dx+\int g(x)\;dx$$
And, likewise, constants ‘go through’ the integral sign:
$$\int c\cdot f(x)\;dx=c\cdot \int f(x)\;dx$$

For example, it is easy to integrate polynomials, even including terms like $\sqrt{x}$ and more general power functions. The only thing to watch out for is terms $x^{-1}={1\over x}$, since these integrate to $\ln x$ instead of a power of $x$. So $$\int 4x^5-3x+11-17\sqrt{x}+{3\over x}\;dx= {4x^6\over 6}-{3x^2\over 2}+11x-{ 17x^{3/2} \over 3/2 }+3\ln x+C$$ Notice that we need to include just one ‘constant of integration’.

Other basic formulas obtained by reversing differentiation formulas: \begin{align*} \int a^x \;dx&= {a^x\over \ln a}+C \\ \int \log_a x\;dx&={1\over \ln a}\cdot{1\over x}+C \\ \int { 1 \over \sqrt{1-x^2 }} \; dx&=\arcsin x+C\\ \int { 1 \over x\sqrt{x^2-1 }} \; dx&=\hbox{ arcsec}\, x+C \end{align*}

Sums of constant multiples of all these functions are easy to integrate: for example, $$\int 5\cdot 2^x-{ 23 \over x\sqrt{x^2-1 }}+5x^2\;dx= {5\cdot 2^x\over \ln 2}-23\,\hbox{arcsec}\,x+{5x^3\over 3}+C$$

#### Exercises

- $\int 4x^3-3\cos x+{ 7 \over x }+2\;dx=?$
- $\int 3x^2+e^{2x}-11+\cos x\,dx=?$
- $\int \sec^2 x\,dx=?$
- $\int { 7 \over 1+x^2 }\; dx=?$
- $\int 16x^7-\sqrt{x}+{ 3 \over \sqrt{x }}\; dx=?$
- $\int 23 \sin x-{ 2 \over \sqrt{1-x^2 }}\; dx=?$