Partial integration

To differentiate a product:

(u⋅v)'=u'⋅v+u⋅v'

By integrating this:

∫(u⋅v)'=∫u'⋅v+∫u⋅v'
∫(u⋅v)'=∫u'⋅v+∫u⋅v'
u⋅v=∫u'⋅v+∫u⋅v'
u⋅v-∫u⋅v'=∫u'⋅v
∫u'⋅v=u⋅v-∫u⋅v'

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 ∫÷{dx}{a⋅x²+b⋅x+c}, a≠0

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

I:=∫÷{dx}{4⋅x²+4⋅x+6}=∫÷{dx}{(2⋅x-1)²+5}={÷{1}{5}}⋅∫÷{dx}{({÷{2⋅x-1}{√5}})²+1}

Substitute u:=÷{2⋅x-1}{√5}.

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:

For ∫÷{dx}{√{a⋅x²+b⋅x+c}}, a≠0

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

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 E:=C-÷{B⋅β}{2}.

J:=∫÷{2⋅x+β}{(x²+βx+γ)^j}dx

Calculate J by substituting u:=x²+β⋅x+γ.

K:=∫÷{dx}{(x²+βx+γ)^j}

The equation x²+βx+γ=0 has no real solutions.

Calculate K by substituting to arrive at:

K=∫÷{du}{(1+u²)^j}, j\geq 0

If j=1, then:

∫÷{du}{(1+u²)^1}=(\arctan u)+C

For j>1, 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 ∫R(x, √{1+x²})dx

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

For ∫R(x, √{x²-1})dx

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

For ∫R(x, √{1-x²})dx

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

For ∫R(x, √{a⋅x²+b⋅x+c})dx

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 -b≤a≤b.