Partial integration

To differentiate a product:

By integrating this:

For $∫ (\ln x) dx$

$∫ (\ln x) dx=∫ 1⋅(\ln x) dx=x⋅(\ln x)-∫ x⋅÷{1}{x} dx$

$∫ (\ln x) dx=x⋅(\ln x)-∫ 1 dx$

$∫ (\ln x) dx=x⋅(\ln x)-x$

$∫ (\ln x) dx=x⋅((\ln x)-1)$

To get the idea, compare differentiation: $(x⋅\ln x)'=x'⋅(\ln x)+x⋅(\ln x)'=1⋅(\ln x)+x⋅÷{1}{x}=(\ln x)+1$ .

For

To calculate, complete the square and substitute, for example:

Substitute .

$I={÷{1}{2⋅√5}}⋅∫÷{du}{u²+1}=÷{1}{2⋅√5}⋅(\arctan u)+C=÷{1}{2⋅√5}⋅(\arctan ÷{2⋅x-1}{√5})+C$

The goal is to rewrite the integral by substitution in order for it to take one of these forms:

$∫÷{du}{1+u²}=(\arctan u)+C$
$∫÷{du}{1-u²}=(\artanh u)+C$

For

To calculate, complete the square and substitute (using a linear substitution) in order for it to take one of these forms:

$∫÷{du}{u}=(\ln u)+C$
$∫÷{du}{√{1+u²}}=(\arsinh u)+C$
$∫÷{du}{√{u²-1}}=(\arcosh u)+C$
$∫÷{du}{√{1-u²}}=(\arcsin u)+C$

For $∫÷{A}{(x-α)^j}dx, j∈N$

$÷{A}{1-j}⋅(x-α)^{1-j}, j>1; \ln |x-α|, j=1$

For $∫÷{B⋅x+C}{(x²+β⋅x+γ)^j}dx, j∈N$

$I:=∫÷{B⋅x+C}{(x²+β⋅x+γ)^j}dx=÷{B}{2}⋅∫÷{2⋅x+β}{(x²+βx+γ)^j}dx+E⋅∫÷{dx}{(x²+βx+γ)^j}$

with .

$J:=∫÷{2⋅x+β}{(x²+βx+γ)^j}dx$

Calculate J by substituting .

$K:=∫÷{dx}{(x²+βx+γ)^j}$

The equation has no real solutions.

Calculate K by substituting to arrive at:

$K=∫÷{du}{(1+u²)^j}, j\geq 0$

If , then:

$∫÷{du}{(1+u²)^1}=(\arctan u)+C$

For , recurse.

Let R(x) be a rational function.

For $∫R(x,(÷{a⋅x+b}{c⋅x+d})^÷{1}{k})⋅dx$

Substitute $t:=({÷{a⋅x+b}{c⋅x+d}})^÷{1}{k}$ to get a rational function in t.

For $∫R(e^{a⋅x})dx$

Substitute $t:=e^{a⋅x}$ to get a rational function in t.

For $∫R(\sin x, \cos x)dx$

Substitute $t:=\tan ÷{x}{2}$ to get a rational function in t.

For

Substitute $x:=\sinh t$ to get an exponential integral.

For

Substitute $x:=\cosh t$ to get an exponential integral.

For

Substitute $x:=\cos t$ or $x:=\sin t$ to get a function with just sin and cos in them.

For

complete the square.

For $∫P(x)⋅e^{a⋅x}dx$ with a polynom P

partial integration to decrease the order of the polynom.

For $∫P(x)⋅\sin(a⋅x) dx$ with a polynom P

partial integration to decrease the order of the polynom.

For $∫P(x)⋅\cos(a⋅x) dx$ with a polynom P

partial integration to decrease the order of the polynom.

Uneigentliche Integrale

Locally Integrable

A function is locally integrable if it is at least integrable in a range [a,x).

One can also do this with the range (x,a].

Limit

By crossing the border, one can define the Uneigentliches Integral like this:

$∫_{a}^{∞} f(x) dx:=\lim_{x\rightarrow ∞} ∫_{a}^{x} f(x) dx$

Multi-dimensional partial integration

Let Ω be a subset of R^n with a piecemeal smooth border ∂Ω. Let the orientation of the border be the normal vector n.

Let v be a differentiable vector field on the environment Ω and φ a steady differentiable scalar field on Ω.

Let the abbreviation dS mean n⋅dS.

Then follows, reminiscent of the product integral rule:

$∫_Ω φ⋅(∇⋅{\bf v}) dV=∫_{∂Ω} φ⋅{\bf v}⋅d{\bf S}-∫_Ω (∇φ)⋅{\bf v} dV$

$∫_{∂Ω} φ⋅{\bf v}⋅d{\bf S}=∫_Ω φ⋅(∇⋅{\bf v}) dV+∫_Ω (∇φ)⋅{\bf v} dV$

Estimating Integrals

$\left|∫_a^b~g(x) dx\right|≤|b-a|⋅\|g\|_∞$

Connection between Integrals and Differentials

$∫_{-b}^{b} X[x]⋅dx=X[a]⋅2⋅b$ where

.