Mathematics

Indefinite Integral. Antiderivatives

Concept of antiderivative, properties of the indefinite integral, basic integration, integration by substitution, integration by parts, and rational functions.

Antiderivative: An antiderivative (or primitive) of a function f(x) is another function F(x) such that F'(x)=f(x)
Indefinite integral: The indefinite integral of a function f(x) is the set F(x) + C of all its antiderivatives. It is denoted by:
∫f(x)dx=F(x)+C\int f(x)dx = F(x) + C
In some cases, it will be possible to reduce the difficulty of the integral by using the linearity property of the integral, breaking the integral down into simpler pieces. Other very useful techniques in some cases will be adding and subtracting the same number, or multiplying and dividing by the same expression. Let's see some examples:
∫(x2+3x−2)dx=∫x2dx+∫3xdx−∫2dx==x33+3x22−2x+C\begin{aligned} \int(x^2 + 3x - 2)dx &= \int x^2 dx + \int 3x dx - \int 2 dx = \\ &= \frac{x^3}{3} + \frac{3x^2}{2} - 2x + C \end{aligned}
∫x2x2+1dx=∫x2+1−1x2+1dx=∫x2+1x2+1dx−∫1x2+1dx==∫dx−∫1x2+1dx=x−arctan⁡x+C\begin{aligned} \int \frac{x^2}{x^2+1}dx &= \int \frac{x^2+1-1}{x^2+1}dx = \int \frac{x^2+1}{x^2+1}dx - \int \frac{1}{x^2+1}dx = \\ &= \int dx - \int \frac{1}{x^2+1}dx = x - \arctan x + C \end{aligned}
∫23x+1dx=13∫3⋅23x+1dx==23∫33x+1dx=23ln⁡∣3x+1∣+C\begin{aligned} \int \frac{2}{3x+1}dx &= \frac{1}{3}\int \frac{3 \cdot 2}{3x+1}dx = \\ &= \frac{2}{3}\int \frac{3}{3x+1}dx = \frac{2}{3}\ln|3x+1| + C \end{aligned}
  • A substitution of a certain expression of x by t (or u) is proposed.
  • Both expressions are differentiated to obtain the differential dx.
  • With the proposed substitution, it must be possible to replace all the elements of the integral so that it becomes simpler.
  • Once the integral is evaluated, we must undo the substitution, replacing t with the corresponding expression of x.
Examples:
∫x1−x2dxLet{1−x2=t2−2x⋅dx=2t⋅dtx⋅dx=−t⋅dt}==∫−tt2dt=−∫dt=−t+C=−1−x2+C\begin{aligned} \int \frac{x}{\sqrt{1-x^2}}dx \quad & \text{Let} \quad \left\{\begin{matrix} 1-x^2 = t^2 \\ -2x \cdot dx = 2t \cdot dt \\ x \cdot dx = -t \cdot dt \end{matrix}\right\} = \\ &= \int \frac{-t}{\sqrt{t^2}}dt = -\int dt = -t + C = -\sqrt{1-x^2} + C \end{aligned}
∫sin⁡4x⋅cos⁡x⋅dxLet{sin⁡x=tcos⁡x⋅dx=dt}==∫t4dt=t55+C=sin⁡5x5+C\begin{aligned} \int \sin^4 x \cdot \cos x \cdot dx \quad & \text{Let} \quad \left\{\begin{matrix} \sin x = t \\ \cos x \cdot dx = dt \end{matrix}\right\} = \\ &= \int t^4 dt = \frac{t^5}{5} + C = \frac{\sin^5 x}{5} + C \end{aligned}
Some common substitutions
The integrand containsSuggested substitution
R(ax)R(a^x)ax=ta^x = t
R(ln⁡x)R(\ln x)ln⁡x=t\ln x = t
R(sin⁡mx,cos⁡nx)R(\sin^m x, \cos^n x)m=odd⇒cos⁡x=tm = odd \Rightarrow \cos x = t
n=odd⇒sin⁡x=tn = odd \Rightarrow \sin x = t
n,m=even⇒tan⁡x=tn, m = even \Rightarrow \tan x = t
other⇒tan⁡(x/2)=tother \Rightarrow \tan(x/2) = t
R(a2+x2)R(\sqrt{a^2+x^2})x=a⋅tan⁡tx = a \cdot \tan t
R(a2−x2)R(\sqrt{a^2-x^2})x=a⋅sin⁡tx = a \cdot \sin t
R(x2−a2)R(\sqrt{x^2-a^2})x=a⋅sec⁡tx = a \cdot \sec t
In some integrals where we have a product of functions u and v, we can use the integration by parts formula:
∫u⋅dv=u⋅v−∫v⋅du\int u \cdot dv = u \cdot v - \int v \cdot du
Examples:
∫x⋅ln⁡x⋅dxLet{ln⁡x=u⇒1xdx=dux⋅dx=dv⇒x22=v}==x22ln⁡x−∫x22⋅1x⋅dx=x22ln⁡x−12∫x⋅dx=x22ln⁡x−x24+C\begin{aligned} \int x \cdot \ln x \cdot dx \quad & \text{Let} \quad \left\{\begin{matrix} \ln x = u \Rightarrow \frac{1}{x}dx = du \\ x \cdot dx = dv \Rightarrow \frac{x^2}{2} = v \end{matrix}\right\} = \\ &= \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} \cdot dx = \frac{x^2}{2} \ln x - \frac{1}{2} \int x \cdot dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C \end{aligned}
∫ln⁡x⋅dxLet{ln⁡x=u⇒1xdx=dudx=dv⇒x=v}==x⋅ln⁡x−∫x⋅1x⋅dx=x⋅ln⁡x−∫dx=x⋅ln⁡x−x+C\begin{aligned} \int \ln x \cdot dx \quad & \text{Let} \quad \left\{\begin{matrix} \ln x = u \Rightarrow \frac{1}{x}dx = du \\ dx = dv \Rightarrow x = v \end{matrix}\right\} = \\ &= x \cdot \ln x - \int x \cdot \frac{1}{x} \cdot dx = x \cdot \ln x - \int dx = x \cdot \ln x - x + C \end{aligned}
To decide which function to choose as u and which as dv, we can follow the LIATE rule. The order of the letters in this word gives us the priority for choosing the function for u.
  • L: Logarithmic
  • I: Inverse trigonometric (arcsin / arccos)
  • A: Algebraic (polynomials)
  • T: Trigonometric (sin / cos)
  • E: Exponential
We will consider integrals of the type P(x)/Q(x)P(x)/Q(x), where the degree of the numerator is strictly less than the degree of the denominator. If the degree of the numerator is greater than or equal to the degree of the denominator, we first decompose the integral by polynomial long division. Ex.:
∫P(x)Q(x)dx=∫C(x)dx+∫R(x)Q(x)dxEx.: ∫2x2+1x2−1dx=∫2dx+∫3x2−1dx\begin{aligned} \int \frac{P(x)}{Q(x)} dx = \int C(x) dx + \int \frac{R(x)}{Q(x)} dx \qquad \text{Ex.: } \int \frac{2x^2+1}{x^2-1} dx = \int 2 dx + \int \frac{3}{x^2-1} dx \end{aligned}
Type 1) Integration of functions of the form: ∫Aax+bdx\displaystyle \int \frac{A}{ax+b} dx. They reduce to a logarithmic integral ∫f′fdx=ln⁡∣f∣+C\displaystyle \int \frac{f'}{f} dx = \ln|f| + C. Ex.:
∫32x−5dx=12∫3⋅22x−5dx=32∫22x−5dx={It is basic}=32ln⁡∣2x−5∣+C\int \frac{3}{2x-5} dx = \frac{1}{2} \int \frac{3 \cdot 2}{2x-5} dx = \frac{3}{2} \int \frac{2}{2x-5} dx = \{\text{It is basic}\} = \frac{3}{2} \ln|2x-5| + C
Type 2) Integration of functions of the form: ∫A(ax+b)ndx\displaystyle \int \frac{A}{(ax+b)^n} dx. They reduce to a power integral ∫fm⋅f′dx=fm+1m+1+C\displaystyle \int f^m \cdot f' dx = \frac{f^{m+1}}{m+1} + C. Ex.:
∫3(2x−5)4dx=12∫3⋅2(2x−5)4dx=32∫2(2x−5)−4dx=={It is basic}=32(2x−5)−3−3+C=−12(2x−5)3+C\begin{aligned} \int \frac{3}{(2x-5)^4} dx &= \frac{1}{2} \int \frac{3 \cdot 2}{(2x-5)^4} dx = \frac{3}{2} \int 2(2x-5)^{-4} dx = \\ &= \{\text{It is basic}\} = \frac{3}{2} \frac{(2x-5)^{-3}}{-3} + C = -\frac{1}{2(2x-5)^3} + C \end{aligned}
Type 3) Integration of functions of the form: ∫Aax2+bx+cdx\displaystyle \int \frac{A}{ax^2+bx+c} dx. Where the polynomial in the denominator has no real roots.
They are transformed into an arctangent-type integral by expressing the denominator as the sum of a perfect square and a constant term. Ex.:
∫32x2+4x+10dx=12∫3x2+2x+5dx=12∫3x2+2x+1+4dx=32∫1(x+1)2+4dx=={It is basic}=32⋅12arctan⁡x+12+C\begin{aligned} \int \frac{3}{2x^2+4x+10} dx &= \frac{1}{2} \int \frac{3}{x^2+2x+5} dx = \frac{1}{2} \int \frac{3}{x^2+2x+1+4} dx = \frac{3}{2} \int \frac{1}{(x+1)^2+4} dx = \\ &= \{\text{It is basic}\} = \frac{3}{2} \cdot \frac{1}{2} \arctan \frac{x+1}{2} + C \end{aligned}
Type 4) Integration of functions of the form: ∫Ax+Bax2+bx+cdx\displaystyle \int \frac{Ax+B}{ax^2+bx+c} dx. Where the polynomial in the denominator has no real roots.
It is decomposed into the sum of a basic logarithmic integral, plus another of type 3, arctangent. Ex.:
∫2x−3x2+x+1dx=∫2x+1−1−3x2+x+1dx=∫2x+1x2+x+1dx+∫−4x2+x+1dx\int \frac{2x-3}{x^2+x+1} dx = \int \frac{2x+1-1-3}{x^2+x+1} dx = \int \frac{2x+1}{x^2+x+1} dx + \int \frac{-4}{x^2+x+1} dx
Of the two resulting integrals, the first one is a basic logarithmic type. The second one is a type 3 integral.
Partial fraction decomposition
This is the reverse process of adding rational expressions. The goal is to transform a rational function into the sum of simpler ones. The first step is to find the roots and factor the denominator. For each factor obtained, we will have the corresponding fractions as shown below:
  • For each distinct linear factor (x−a)(x-a), we will have a term of the form Ax−a\displaystyle \frac{A}{x-a}, which results in a type 1 integral
  • For each repeated linear factor (x−b)n(x-b)^n, we will have nn terms; B1x−b+B2(x−b)2+…+Bn(x−b)n\displaystyle \frac{B_1}{x-b} + \frac{B_2}{(x-b)^2} + \ldots + \frac{B_n}{(x-b)^n}, which result in type 1 and type 2 integrals
  • For each irreducible quadratic factor (ax2+bx+c)(ax^2+bx+c), we will have a term of the form Cx+Dax2+bx+c\displaystyle \frac{Cx+D}{ax^2+bx+c}, which is a type 4 integral
Example:
7x+1(x−1)(x+1)2=Ax−1+Bx+1+C(x+1)2\frac{7x+1}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}
To determine the unknowns A, B, and C, we must equate the numerators of both sides:
7x+1=A(x+1)2+B(x+1)(x−1)+C(x−1)7x+1 = A(x+1)^2 + B(x+1)(x-1) + C(x-1)
We substitute as many arbitrary values for xx as we have unknowns, to obtain as many equations:
x=1⇒8=4A⇒A=2x=−1⇒−6=−2C⇒C=3x=0⇒1=A−B−C⇒B=−2\begin{aligned} x=1 \quad &\Rightarrow \quad 8 = 4A \quad &\Rightarrow \quad &A=2 \\ x=-1 \quad &\Rightarrow \quad -6 = -2C \quad &\Rightarrow \quad &C=3 \\ x=0 \quad &\Rightarrow \quad 1 = A-B-C \quad &\Rightarrow \quad &B=-2 \end{aligned}
We obtain:
2x−1−2x+1+3(x+1)2\frac{2}{x-1} - \frac{2}{x+1} + \frac{3}{(x+1)^2}