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