# Intefinite integrals

Source: algebrica.org — CC BY-NC 4.0
https://algebrica.org/indefinite-integrals/

## Primitives

Differentiation assigns to each function a unique [derivative](../derivatives.md) by definition. The inverse process asks whether, for a given function \\(f(x)\\), there exists a function \\(F(x)\\) whose derivative is exactly \\(f(x)\\). Such a function is called a primitive of \\(f\\). Formally, \\(F(x)\\) is a primitive of \\(f(x)\\) on the [interval](../intervals.md) \\([a, b]\\) if \\(F\\) is differentiable throughout \\([a, b]\\) and:

\\[F'(x) = f(x), \quad \forall x \in [a, b]\\]

Not every function admits a primitive on a given interval. A sufficient condition is [continuity](../continuous-functions.md): every continuous function on a closed interval \\([a, b]\\) admits a primitive there. The converse does not hold in general. As an example, if \\(f(x) = 3x^2\\), a primitive is \\(F(x) = x^3\\), since:

\\[\frac{d}{dx} x^3 = 3x^2\\]

Unlike derivatives, primitives are not unique. Since the derivative of any constant is zero, the functions \\(x^3\\), \\(x^3 + 5\\), and \\(x^3 - \frac{1}{2}\\) are all primitives of \\(3x^2\\). More generally, if \\(F(x)\\) is a primitive of \\(f(x)\\), then so is \\(F(x) + c\\) for any \\(c \in \mathbb{R}\\), since:

\\[\frac{d}{dx}[F(x) + c] = F'(x) = f(x)\\]

Conversely, any two primitives of the same function differ by a constant. If \\(F_1(x)\\) and \\(F_2(x)\\) are both primitives of \\(f(x)\\), then:

\\[\frac{d}{dx}[F_1(x) - F_2(x)] = F\_1'(x) - F\_2'(x) = f(x) - f(x) = 0\\]

which implies \\(F_1(x) - F_2(x) = c\\) for some constant \\(c \in \mathbb{R}\\).

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## What is the indefinite integral

The indefinite integral of a function \\(f(x)\\) is the set of all its primitives. Since any two primitives differ by a constant, the entire family is expressed as \\(F(x) + c\\) for \\(c \in \mathbb{R}\\), and is denoted:

\\[\int f(x) \\, dx = F(x) + c, \quad c \in \mathbb{R}\\]

From this definition it follows directly that:

\\[\frac{d}{dx}\left[ \int f(x) \\, dx \right] = f(x)\\]

Differentiating an indefinite integral returns the original function. This relationship is made precise by the [Fundamental Theorem of Calculus](../fundamental-theorem-of-calculus.md), which establishes the formal connection between differentiation and integration.

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## Example 1

Find the primitive of \\(f(x) = 3x\\) that passes through the point \\((2, 1)\\). The general primitive is obtained by integrating:

\\[F(x) = \int 3x \\, dx = \frac{3}{2}x^2 + c\\]

To determine the constant, the condition \\(F(2) = 1\\) is imposed:

\\[\begin{align}
\frac{3}{2}(2)^2 + c &= 1 \\\\[6pt]
6 + c &= 1 \\\\[6pt]
c &= -5
\end{align}\\]

The unique primitive satisfying the given a is \\(F(x) = \frac{3}{2}x^2 - 5\\).

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## Linearity properties

The indefinite integral is a linear operator. The integral of a sum of integrable functions equals the sum of their integrals:

\\[\int \left[ f(x) + g(x) \right] \\, dx = \int f(x) \\, dx + \int g(x) \\, dx \tag{1}\\]

A constant factor can be moved outside the integral sign:

\\[\int k f(x) \\, dx = k \int f(x) \\, dx, \quad \forall k \in \mathbb{R} \tag{2}\\]

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## Example 2

Compute the integral of \\(f(x) = 3x^2 + 2x\\). Applying property \\(1\\), the integral splits into two terms, each of which falls under the power rule:

\\[\int (3x^2 + 2x) \\, dx = \int 3x^2 \\, dx + \int 2x \\, dx\\]

The two integration constants arising from each term combine into a single arbitrary constant, giving \\(x^3 + x^2 + c\\) with \\(c \in \mathbb{R}\\).

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## Example 3

Compute the integral of \\(f(x) = 5\sin(x)\\). Applying property \\(2\\), the constant factor is moved outside the integral:

\\[\int 5\sin(x) \\, dx = 5 \int \sin(x) \\, dx\\]

Since the integral of \\(\sin(x)\\) is \\(-\cos(x)\\), the result is \\(-5\cos(x) + c\\) with \\(c \in \mathbb{R}\\).

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## Integral of a power function

The integral of a power function \\(x^a\\) with \\(a \in \mathbb{R}\\) and \\(a \neq -1\\) is given by:

\\[\int x^a \\, dx = \frac{x^{a+1}}{a+1} + c\\]

The case \\(a = -1\\) requires a separate treatment and is discussed in the next section.

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## Example 4

Compute the following integral:

\\[\int (3x^4 + 5x^2) \\, dx\\]

Applying linearity, the constant factors are moved outside and the power rule is applied to each term:

\\[\int (3x^4 + 5x^2) \\, dx = 3 \int x^4 \\, dx + 5 \int x^2 \\, dx = 3 \cdot \frac{x^5}{5} + 5 \cdot \frac{x^3}{3} + c\\]

The result is \\(\dfrac{3}{5}x^5 + \dfrac{5}{3}x^3 + c\\) with \\(c \in \mathbb{R}\\).

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## Example 5

Compute the following integral:

\\[\int \left( 4x^3 - \frac{3}{\sqrt{x}} + 2\cos x \right) \\, dx\\]

Applying linearity, the integral splits into three terms:

\\[\int 4x^3 \\, dx - \int 3x^{-1/2} \\, dx + \int 2\cos x \\, dx\\]

The first term follows directly from the power rule: \\(\int 4x^3 \\, dx = x^4\\). For the second, rewriting \\(\frac{1}{\sqrt{x}}\\) as \\(x^{-1/2}\\) and applying the power rule with \\(a = -\frac{1}{2}\\) gives \\(\int 3x^{-1/2} \\, dx = 6\sqrt{x}\\). The third term follows from the standard integral of the cosine: \\(\int 2\cos x \\, dx = 2\sin x\\).

Assembling the three contributions:

\\[\int \left( 4x^3 - \frac{3}{\sqrt{x}} + 2\cos x \right) \\, dx = x^4 - 6\sqrt{x} + 2\sin x + c\\]

> The result can be verified by differentiating \\(x^4 - 6\sqrt{x} + 2\sin x + c\\) term by term, which returns the original integrand.

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## The logarithmic integral

When \\(a = -1\\), the power rule formula produces a zero denominator and does not apply. The integral in this case is:

\\[\int \frac{1}{x} \\, dx = \ln |x| + c\\]

This follows from the fact that \\(\frac{d}{dx} \ln |x| = \frac{1}{x}\\), which holds for all \\(x \neq 0\\). The absolute value is necessary because \\(\ln\\) is defined only for positive arguments, while \\(\frac{1}{x}\\) is defined on both \\((-\infty, 0)\\) and \\((0, +\infty)\\).

> The identity \\(\int \frac{1}{x} \\, dx = \ln |x| + c\\) holds separately on \\((-\infty, 0)\\) and on \\((0, +\infty)\\). On each interval the arbitrary constant may take a different value, so the most general antiderivative of \\(\frac{1}{x}\\) on its full domain is not a single expression \\(\ln |x| + c\\) with one constant, but a piecewise family with independent constants on the two components.

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## Fundamental integration rules

|                  |                                                                                  |
| ---------------- | -------------------------------------------------------------------------------- |
| Linearity        | \\[ \int \left( f(x) + g(x)\right)\\, dx = \int f(x)\\, dx + \int g(x)\\, dx \\] |
| Linearity        | \\[ \int k\\, f(x)\\, dx = k \int f(x)\\, dx \\]                                 |
| Power rule       | \\[ \int x^a\\, dx = \dfrac{x^{a+1}}{a+1} + c \quad a \neq -1 \\]                |
| Logarithmic case | \\[ \int \dfrac{1}{x} \\,dx = \ln\|x\| + c \\]                                   |
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## Common Integrals

Below is a summary of the most common basic integrals, useful in calculus and for transforming complex expressions into simpler, well-known forms.

+ \\[\int \frac{1}{x} \\, dx = \ln |x| + c\\] [Further reading](../integral-of-rational-functions.md)

+ \\[\int a^x \\, dx = \frac{1}{\ln a} \cdot a^x + c\\] [Further reading](../integral-of-the-exponential-function.md)

+ \\[\int \sin x \\, dx = -\cos x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \cos x \\, dx = \sin x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \frac{1}{\sin^2x}\\, dx  = \cot x +c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \frac{1}{\cos^2x}\\, dx  = \tan x +c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \sec^2 x \\, dx = \tan x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \sec x \tan x \\, dx = \sec x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \csc^2 x \\, dx = -\cot x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \csc x \cot x \\, dx = -\csc x + c\\] [Further reading](../integral-of-trigonometric-functions.md)
+ \\[\int \frac{dx}{1 + x^2} = \arctan x + c\\]
+ \\[\int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + c\\]

> These identities hold on any interval where the integrand is defined and continuous.