Calculus

Integration of Functions

Integration of Functions Logo

The Indefinite Integral and Basic Rules of Integration

Antiderivatives and the Indefinite Integral

Let a function f (x) be defined on some interval I. The function F (x) is called an antiderivative of f (x), if

\[F^\prime\left( x \right) = f\left( x \right)\]

for all x in the interval I.

There is an infinite number of antiderivatives of a function f (x), all differing only by a constant C:

\[\left( {F\left( x \right) + C} \right)^\prime = F^\prime\left( x \right) + C^\prime = f\left( x \right) + 0 = f\left( x \right).\]

The set of all antiderivatives for a function f (x) is called the indefinite integral of f (x) and is denoted as

\[{\int} {{f\left( x \right)}{dx}} = F\left( x \right) + C,\;\;\text{if}\;\;F^\prime\left( x \right) = f\left( x \right).\]

In this definition, the ∫ is called the integral symbol, f (x) is called the integrand, x is called the variable of integration, dx is called the differential of the variable x, and C is called the constant of integration.

Indefinite Integral of Some Common Functions

Integration is the reverse process of differentiation, so the table of basic integrals follows from the table of derivatives.

It is supposed here that \(a,\) \(p\left( {p \ne 1} \right),\) \(C\) are real constants, \(b\) is the base of the exponential function \(\left( {b \ne 1, b \gt 0} \right).\)

Table of integrals

Properties of the Indefinite Integral

  1. If \(a\) is some constant, then
    \[\int {af\left( x \right)dx} = a\int {f\left( x \right)dx},\]
    i.e. the constant coefficient can be carried outside the integral sign.
  2. For functions \(f\left( x \right)\) and \(g\left( x \right),\)
    \[\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} ,\]
    i.e. the indefinite integral of the sum (difference) equals to the sum (difference) of the integrals.

Calculation of integrals using the linear properties of indefinite integrals and the table of basic integrals is called direct integration.

Solved Problems

Example 1.

Evaluate the indefinite integral \[\int {\left( {3{x^2} - 6x + 2\cos x} \right)dx} .\]

Solution.

Applying the properties \(1\) and \(2,\) we have

\[I = \int {\left( {3{x^2} - 6x + 2\cos x} \right)dx} = \int {3{x^2}dx} - \int {6xdx} + \int {2\cos xdx} = 3{\int {{x^2}dx}} - 6{\int {xdx}} + 2{\int {\cos xdx}} .\]

All three integrals can be evaluated using the integration table. This yields:

\[I = 3 \cdot {\frac{{{x^3}}}{3}} - 6 \cdot {\frac{{{x^2}}}{2}} + 2 \cdot {\sin x} + C = {x^3} - 3{x^2} + 2\sin x + C.\]

Example 2.

Find the indefinite integral \[\int {\left( {1 + x} \right)\left( {1 + 2x} \right)dx}.\]

Solution.

We can simplify the integrand:

\[\left( {1 + x} \right)\left( {1 + 2x} \right) = 1 + x + 2x + 2{x^2} = 2{x^2} + 3x + 1.\]

Then the integral is given by

\[\int {\left( {1 + x} \right)\left( {1 + 2x} \right)dx} = \int {\left( {2{x^2} + 3x + 1} \right)dx} = \int {2{x^2}dx} + \int {3xdx} + \int {1dx} = 2\int {{x^2}dx} + 3\int {xdx} + \int {dx} = 2 \cdot \frac{{{x^3}}}{3} + 3 \cdot \frac{{{x^2}}}{2} + x + C = \frac{{2{x^3}}}{3} + \frac{{3{x^2}}}{2} + x + C.\]

Example 3.

Find the indefinite integral \[\int {\left( {\frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}} \right)dx}.\]

Solution.

By the sum rule,

\[I = \int {\left( {\frac{1}{{{x^2}}} - \frac{1}{{{x^3}}}} \right)dx} = \int {\frac{{dx}}{{{x^2}}}} - \int {\frac{{dx}}{{{x^3}}}} .\]

The integrands in both integrals are power functions, so we have

\[I = \int {{x^{ - 2}}dx} - \int {{x^{ - 3}}dx} = \frac{{{x^{ - 1}}}}{{\left( { - 1} \right)}} - \frac{{{x^{ - 2}}}}{{\left( { - 2} \right)}} + C = - \frac{1}{x} + \frac{1}{{2{x^2}}} + C.\]

Example 4.

Calculate \[\int {\left( {\sqrt x + \sqrt[3]{x}} \right)dx}.\]

Solution.

\[\int {\left( {\sqrt x + \sqrt[3]{x}} \right)dx} = \int {\sqrt x dx} + \int {\sqrt[3]{x}dx} = \int {{x^{\frac{1}{2}}}dx} + \int {{x^{\frac{1}{3}}}dx} = \frac{{{x^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + \frac{{{x^{\frac{1}{3} + 1}}}}{{\frac{1}{3} + 1}} + C = \frac{{2{x^{\frac{3}{2}}}}}{3} + \frac{{3{x^{\frac{4}{3}}}}}{4} = \frac{{2\sqrt {{x^3}} }}{3} + \frac{{3\sqrt[3]{{{x^4}}}}}{4} + C.\]

Example 5.

Find the indefinite integral \[\int {\frac{{x + 1}}{{\sqrt x }} dx}.\]

Solution.

We write the integrals as the sum of two integrals and calculate them separately:

\[\int {\frac{{x + 1}}{{\sqrt x }}dx} = \int {\left( {\frac{x}{{\sqrt x }} + \frac{1}{{\sqrt x }}} \right)dx} = \int {\left( {\sqrt x + \frac{1}{{\sqrt x }}} \right)dx} = \int {\sqrt x dx} + \int {\frac{{dx}}{{\sqrt x }}} = \frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} + 2\sqrt x + C = \frac{{2\sqrt {{x^3}} }}{3} + 2\sqrt x + C.\]

Example 6.

Find the indefinite integral \[\int {{{\left( {x + \sqrt x } \right)}^2}dx}.\]

Solution.

Expand the square in the integrand:

\[I = \int {{{\left( {x + \sqrt x } \right)}^2}dx} = \int {\left( {{x^2} + 2x\sqrt x + {{\left( {\sqrt x } \right)}^2}} \right)dx} = \int {\left( {{x^2} + 2{x^{\frac{3}{2}}} + x} \right)dx} .\]

Using the basic properties of integrals, we have

\[I = \int {\left( {{x^2} + 2{x^{\frac{3}{2}}} + x} \right)dx} = \int {{x^2}dx} + 2\int {{x^{\frac{3}{2}}}dx} + \int {xdx} .\]

The last expression contains only table integrals. Then

\[I = \frac{{{x^3}}}{3} + 2 \cdot \frac{{{x^{\frac{5}{2}}}}}{{\frac{5}{2}}} + \frac{{{x^2}}}{2} + C = \frac{{{x^3}}}{3} + \frac{{4{x^{\frac{5}{2}}}}}{5} + \frac{{{x^2}}}{2} + C = \frac{{{x^3}}}{3} + \frac{{4\sqrt {{x^5}} }}{5} + \frac{{{x^2}}}{2} + C.\]

Example 7.

Calculate the integral \[\int {\left( {\frac{3}{{\sqrt[3]{x}}} + \frac{2}{{\sqrt x }}} \right)dx}.\]

Solution.

Using the power rule for integrals, we have

\[\int {\left( {\frac{3}{{\sqrt[3]{x}}} + \frac{2}{{\sqrt x }}} \right)dx} = \int {\frac{{3dx}}{{\sqrt[3]{x}}}} + \int {\frac{{2dx}}{{\sqrt x }}} = 3\int {{x^{ - \frac{1}{3}}}dx} + 2\int {{x^{ - \frac{1}{2}}}dx} = 3 \cdot \frac{{{x^{ - \frac{1}{3} + 1}}}}{{ - \frac{1}{3} + 1}} + 2 \cdot \frac{{{x^{ - \frac{1}{2} + 1}}}}{{ - \frac{1}{2} + 1}} + C = \frac{{9{x^{\frac{2}{3}}}}}{2} + 4{x^{\frac{1}{2}}} + C = \frac{{9\sqrt[3]{{{x^2}}}}}{2} + 4\sqrt x + C.\]

Example 8.

Find the indefinite integral \[\int {\left( {\sqrt[3]{x} + {e^3}} \right)dx}.\]

Solution.

Using the basic properties of the integrals, we can write:

\[I = \int {\left( {\sqrt[3]{x} + {e^3}} \right)dx} = \int {\left( {{x^{\frac{1}{3}}} + {e^3}} \right)dx} = \int {{x^{\frac{1}{3}}}dx} + \int {{e^3}dx} = \int {{x^{\frac{1}{3}}}dx} + {e^3}\int {dx} .\]

As you can see, we have table integrals. In the second integral, \({e^3}\) is a constant, so it can be pulled through the integral sign. This yields:

\[I = \int {{x^{\frac{1}{3}}}dx} + {e^3}\int {dx} = \frac{{{x^{\frac{4}{3}}}}}{{\frac{4}{3}}} + {e^3}x + C = \frac{{3\sqrt[3]{{{x^4}}}}}{4} + {e^3}x + C.\]

See more problems on Page 2.

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