Calculus

Integration of Functions

The Indefinite Integral and Basic Formulas of Integration. Table of Integrals.

Page 1
Problems 1-2
Page 2
Problems 3-6

Definition of the Antiderivative and Indefinite Integral

The function \(F\left( x \right)\) is called an antiderivative of \(f\left( x \right),\) if
\[F’\left( x \right) = f\left( x \right).\] The family of all antiderivatives of a function \(f\left( x \right)\) is called the indefinite integral of the function \(f\left( x \right)\) and is denoted by
\[\int {f\left( x \right)dx} .\] Thus, if \(F\) is a particular antiderivative, we may write
\[{\int {f\left( x \right)dx} }={ F\left( x \right) + C,}\] where \(C\) is an arbitrary constant.

Properties of the Indefinite Integral

In formulas given below \(f\) and \(g\) are functions of the variable \(x,\) \(F\) is an antiderivative of \(f,\) and \(a, k, C\) are constants.

  • \({\int {\left[ {f\left( x \right) + g\left( x \right)} \right]dx} }\) \(={ \int {f\left( x \right)dx} }+{ \int {g\left( x \right)dx} }\)
  • \({\int {kf\left( x \right)dx} }={ k\int {f\left( x \right)dx} }\)
  • \({\int {f\left( {ax} \right)dx} }\) \(={ {\large\frac{1}{a}\normalsize} F\left( {ax} \right) + C}\)
  • \({\int {f\left( {ax + b} \right)dx} }\) \(={ {\large\frac{1}{a}\normalsize} F\left( {ax + b} \right) + C}\)

Table of Integrals

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

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

Solved Problems

Click on problem description to see solution.

 Example 1

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

 Example 2

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

 Example 3

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

 Example 4

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

 Example 5

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

 Example 6

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

Example 1.

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 2.

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

Solution.

Transforming the integrand and using the formula for integral of the power function, we have
\[
{\int {\left( {\frac{3}{{\sqrt[\large 3\normalsize]{x}}} + \frac{2}{{\sqrt x }}} \right)dx} }
= {\int {\frac{{3dx}}{{\sqrt[\large 3\normalsize]{x}}}} + \int {\frac{{2dx}}{{\sqrt x }}} }
= {3\int {{x^{ – \large\frac{1}{3}\normalsize}}dx} + 2\int {{x^{ – \large\frac{1}{2}\normalsize}}dx} }
= {3 \cdot \frac{{{x^{ – \large\frac{1}{3}\normalsize + 1}}}}{{ – \large\frac{1}{3}\normalsize + 1}} }+{ 2 \cdot \frac{{{x^{ – \large\frac{1}{2}\normalsize + 1}}}}{{ – \frac{1}{2} + 1}} }+{ C }
= {\frac{{9{x^{\large\frac{2}{3}\normalsize}}}}{2} + 4{x^{\large\frac{1}{2}\normalsize}} + C }
= {\frac{{9\sqrt[\large 3\normalsize]{{{x^2}}}}}{2} + 4\sqrt x + C.}
\]

Page 1
Problems 1-2
Page 2
Problems 3-6