Calculus

Integration of Functions

Integration of Functions Logo

The Indefinite Integral and Basic Rules of Integration

  • Antiderivatives and the Indefinite Integral

    Let a function \(f\left( x \right)\) be defined on some interval \(I.\) The function \(F\left( x \right)\) is called an antiderivative of \(f\left( x \right),\) 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 \(f\left( x \right),\) 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 \(f\left( x \right)\) is called the indefinite integral of \(f\left( x \right)\) and is denoted as

    \[{{\int} {{f\left( x \right)}{dx}} }={ F\left( x \right) + C,\;\;}\kern0pt{\text{if}\;\;F^\prime\left( x \right) = f\left( x \right).}\]

    In this definition, the \(\int {} \) is called the integral symbol, \(f\left( x \right)\) 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.

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

    \(\int {adx} = ax + C\) \(\int {xdx} = {\large\frac{{{x^2}}}{2}\normalsize} + C\)
    \(\int {{x^2}dx} = {\large\frac{{{x^3}}}{3}\normalsize} + C\) \(\int {{x^p}dx} = {\large\frac{{{x^{p + 1}}}}{{p + 1}}\normalsize} + C\)
    \(\int {\large\frac{{dx}}{x}\normalsize} = \ln \left| x \right| + C\) \(\int {{e^x}dx} = {e^x} + C\)
    \(\int {{b^x}dx} = {\large\frac{{{b^x}}}{{\ln b}}\normalsize} + C\) \(\int {\sin xdx} = – \cos x + C\)
    \(\int {\cos xdx} = \sin x + C\) \(\int {\tan xdx} = – {\ln \left| {\cos x} \right|} + C\)
    \(\int {\cot xdx} = {\ln \left| {\sin x} \right|} + C\) \(\int {\sec xdx} = {\ln \left| {\tan\left( {\large\frac{x}{2}\normalsize} + {\large\frac{\pi }{4}\normalsize} \right)} \right| + C} = {\ln \left| {\sec x + \tan x} \right| + C}\)
    \(\int {\csc xdx} = {\ln \left| {\tan\large\frac{x}{2}\normalsize} \right| + C} = {-\ln \left| {\csc x + \cot x} \right| + C}\) \(\int {{\sec^2}xdx} = \tan x + C\)
    \(\int {{\csc^2}xdx} = -\cot x + C\) \(\int {\sec x\tan xdx} = \sec x + C\)
    \(\int {\csc x\cot xdx} = -\csc x + C\) \(\int {\large\frac{{dx}}{{1 + {x^2}}}\normalsize} = \arctan x + C\)
    \(\int {\large\frac{{dx}}{{{a^2} + {x^2}}}\normalsize} = {\large\frac{1}{a}\normalsize}\arctan {\large\frac{x}{a}\normalsize} + C\) \(\int {\large\frac{{dx}}{{1 – {x^2}}}\normalsize} = {\large\frac{1}{2}\normalsize}{\ln \left| {{\large\frac{{1 + x}}{{1 – x}}\normalsize}} \right|} + C\)
    \(\int {\large\frac{{dx}}{{{a^2} – {x^2}}}\normalsize} = {\large\frac{1}{{2a}}\normalsize}\ln\left| {\large{\frac{{a + x}}{{a – x}}\normalsize}} \right| + C\) \(\int {\large\frac{{dx}}{{\sqrt {1 – {x^2}} }}\normalsize} = \arcsin x + C\)
    \(\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize} = \arcsin {\large\frac{x}{a}\normalsize} + C\) \(\int {\large\frac{{dx}}{{\sqrt {{x^2} \pm {a^2}} }}\normalsize} = {\ln \left| {x + \sqrt {{x^2} \pm {a^2}} } \right|} + C\)
    \(\int {\large\frac{{dx}}{{x\sqrt {{x^2} – 1} }}\normalsize} = {\text{arcsec}\left| x \right|} + C\) \(\int {\sinh xdx} = \cosh x + C\)
    \(\int {\cosh xdx} = \sinh x + C\) \(\int {{\text{sech}^2}xdx} = \tanh x + C\)
    \(\int {{\text{csch}^2}xdx} = -\text{coth}\,x + C\) \(\int {\text{sech}\,x\tanh xdx} = – {\text{sech}\,x} + C\)
    \(\int {\text{csch}\,x\coth xdx} = – {\text{csch}\,x} + C\) \(\int {\tanh xdx} = {\ln \cosh x} + C\)
    \(\int {adx} = ax + C\)
    \(\int {xdx} = {\large\frac{{{x^2}}}{2}\normalsize} + C\)
    \(\int {{x^2}dx} = {\large\frac{{{x^3}}}{3}\normalsize} + C\)
    \(\int {{x^p}dx} = {\large\frac{{{x^{p + 1}}}}{{p + 1}}\normalsize} + C\)
    \(\int {\large\frac{{dx}}{x}\normalsize} = {\ln \left| x \right|} + C\)
    \(\int {{e^x}dx} = {e^x} + C\)
    \(\int {{b^x}dx} = {\large\frac{{{b^x}}}{{\ln b}}\normalsize} + C\)
    \(\int {\sin xdx} = – \cos x + C\)
    \(\int {\cos xdx} = \sin x + C\)
    \(\int {\tan xdx} = – {\ln \left| {\cos x} \right|} + C\)
    \(\int {\cot xdx} = {\ln \left| {\sin x} \right|} + C\)
    \(\int {\sec xdx} = {\ln \left| {\tan\left( {\large\frac{x}{2}\normalsize} + {\large\frac{\pi }{4}\normalsize} \right)} \right| + C} = {\ln \left| {\sec x + \tan x} \right| + C}\)
    \(\int {\csc xdx} = {\ln \left| {\tan\large\frac{x}{2}\normalsize} \right| + C} = {-\ln \left| {\csc x + \cot x} \right| + C}\)
    \(\int {{\sec^2}xdx} = \tan x + C\)
    \(\int {{\csc^2}xdx} = -\cot x + C\)
    \(\int {\sec x\tan xdx} = \sec x + C\)
    \(\int {\csc x\cot xdx} = -\csc x + C\)
    \(\int {\large\frac{{dx}}{{1 + {x^2}}}\normalsize} = \arctan x + C\)
    \(\int {\large\frac{{dx}}{{{a^2} + {x^2}}}\normalsize} = {\large\frac{1}{a}\normalsize}\arctan {\large\frac{x}{a}\normalsize} + C\)
    \(\int {\large\frac{{dx}}{{1 – {x^2}}}\normalsize} = {\large\frac{1}{2}\normalsize}\ln \left| {{\large\frac{{1 + x}}{{1 – x}}\normalsize}} \right| + C\)
    \(\int {\large\frac{{dx}}{{{a^2} – {x^2}}}\normalsize} = {\large\frac{1}{{2a}}\normalsize}\ln\left| {\large{\frac{{a + x}}{{a – x}}\normalsize}} \right| + C\)
    \(\int {\large\frac{{dx}}{{\sqrt {1 – {x^2}} }}\normalsize} = \arcsin x + C\)
    \(\int {\large\frac{{dx}}{{\sqrt {{a^2} – {x^2}} }}\normalsize} = \arcsin {\large\frac{x}{a}\normalsize} + C\)
    \(\int {\large\frac{{dx}}{{\sqrt {{x^2} \pm {a^2}} }}\normalsize} = {\ln \left| {x + \sqrt {{x^2} \pm {a^2}} } \right|} + C\)
    \(\int {\large\frac{{dx}}{{x\sqrt {{x^2} – 1} }}\normalsize} = \text{arcsec}\left| x \right| + C\)
    \(\int {\sinh xdx} = \cosh x + C\)
    \(\int {\cosh xdx} = \sinh x + C\)
    \(\int {{\text{sech}^2}xdx} = \tanh x + C\)
    \(\int {{\text{csch}^2}xdx} = -\text{coth}\,x + C\)
    \(\int {\text{sech}\,x\tanh xdx} = – \text{sech}\,x + C\)
    \(\int {\text{csch}\,x\coth xdx} = – \text{csch}\,x + C\)
    \(\int {\tanh xdx} = \ln \cosh x + C\)

    Properties of the Indefinite Integral

    1. If \(a\) is some constant, then \[\cssId{element11}{\int {af\left( x \right)dx} }=\cssId{element12}{ a\int {f\left( x \right)dx},}\] i.e. the constant coefficient can be carried outside the integral sign.
    2. For functions \(f\left( x \right)\) and \(g\left( x \right),\) \[\cssId{element13}{\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} }=\cssId{element14}{ \int {f\left( x \right)dx} }\pm \cssId{element15}{\int {g\left( x \right)dx} ,}\] 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

    Click a problem to see the solution.

    Example 1

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

    Example 2

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

    Example 3

    Find the indefinite integral \(\int {\left( {\large{\frac{1}{{{x^2}}}}\normalsize – \large{\frac{1}{{{x^3}}}}\normalsize} \right)dx}.\)

    Example 4

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

    Example 5

    Find the indefinite integral \(\int {\large{\frac{{x + 1}}{{\sqrt x }}}\normalsize dx}.\)

    Example 6

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

    Example 7

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

    Example 8

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

    Example 9

    Calculate \(\int {\large\frac{{4dx}}{{2 + 3{x^2}}}\normalsize}.\)

    Example 10

    Find the indefinite integral \(\int {\large{\frac{{{x^2}}}{{1 + {x^2}}}}\normalsize dx}.\)

    Example 11

    Evaluate the integral \(\int {\large{\frac{{dx}}{{1 + 2{x^2}}}}\normalsize}.\)

    Example 12

    Find the integral \(\int {\large\frac{{\pi dx}}{{\sqrt {\pi – {x^2}} }}\normalsize}.\)

    Example 13

    Calculate the integral \(\int {\left( {2\cos x – 5\sin x} \right)dx}.\)

    Example 14

    Evaluate the integral \(\int {\large{\frac{{dx}}{{\sqrt {1 – \large{\frac{{{x^2}}}{2}}\normalsize} }}}\normalsize}.\)

    Example 15

    Calculate the integral \(\int {{{\tan }^2}xdx}.\)

    Example 16

    Calculate the integral \(\int {{{\cot }^2}xdx}.\)

    Example 17

    Find the integral \(\int {\large\frac{{dx}}{{{\sin^2}2x}}\normalsize}\) without using a substitution.

    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 \(\int {\left( {\large{\frac{1}{{{x^2}}}}\normalsize – \large{\frac{1}{{{x^3}}}}\normalsize} \right)dx} .\)

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

    Solution.

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

    Example 5.

    Find the indefinite integral \(\int {\large{\frac{{x + 1}}{{\sqrt x }}}\normalsize dx}.\)

    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

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

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    Problems 1-6
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    Problems 7-17