Calculus

Integration of Functions

Integration of Functions Logo

Integration of Irrational Functions

  • To integrate an irrational function containing a term \({x^{\large\frac{m}{n}\normalsize}}\) we make the substitution \(u = {x^{\large\frac{1}{n}\normalsize}}.\)

    To integrate an irrational function involving more than one rational power of \(x\), make a substitution of the form \(u = {x^{\large\frac{1}{n}\normalsize}},\) where \(n\) is chosen to be the least common multiple of the denominators of all the fractional powers that appear in the function.

    An expression of the form \(\sqrt[\large n\normalsize]{{\large\frac{{ax + b}}{{cx + d}}\normalsize}}\) is integrated by the substitution \(u = {\left( {\large\frac{{ax + b}}{{cx + d}}\normalsize} \right)^{\large\frac{1}{n}\normalsize}}.\)

    See Trigonometric and Hyperbolic Substitutions about integration of irrational functions involving \(\sqrt {{a^2} – {x^2}},\) \(\sqrt {{a^2} + {x^2}}\) and \(\sqrt {{x^2} – {a^2}}.\)


    Solved Problems

    Click a problem to see the solution.

    Example 1

    Find the integral \({\large\int\normalsize} {{\large\frac{{\sqrt {x + 9} }}{x}\normalsize} dx}.\)

    Example 2

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

    Example 3

    Evaluate the integral \({\large\int\normalsize} {\large\frac{{dx}}{{x + \sqrt[3]{x}}}\normalsize}.\)

    Example 4

    Evaluate the integral \(\int {\large\frac{{dx}}{{\sqrt[5]{x} – 1}}\normalsize}.\)

    Example 5

    Find the integral \({\large\int\normalsize} {\large\frac{{dx}}{{\sqrt[3]{x} – \sqrt[4]{x}}}\normalsize}.\)

    Example 6

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

    Example 7

    Calculate the integral \({\large\int\normalsize} {\sqrt {{e^x} + 1}\,dx}.\)

    Example 1.

    Find the integral \({\large\int\normalsize} {{\large\frac{{\sqrt {x + 9} }}{x}\normalsize} dx}.\)

    Solution.

    We make the substitution:

    \[
    {u = {\left( {x + 9} \right)^{\large\frac{1}{2}\normalsize}},\;\;}\Rightarrow
    {x + 9 = {u^2},\;\;}\Rightarrow
    {x = {u^2} – 9,\;\;\;}\kern-0.3pt{dx = 2udu.}
    \]

    Then

    \[
    {\int {\frac{{\sqrt {x + 9} }}{x}dx} }
    = {\int {\frac{u}{{{u^2} – 9}} \cdot 2udu} }
    = {2\int {\frac{{{u^2}}}{{{u^2} – 9}}du} }
    = {2\int {\frac{{{u^2} – 9 + 9}}{{{u^2} – 9}}du} }
    = {2\int {\left( {1 + \frac{9}{{{u^2} – 9}}} \right)du} }
    = {2\int {du} + 18\int {\frac{{du}}{{{u^2} – {3^2}}}} }
    = {{2u + 18 \cdot \frac{1}{6}\ln \left| {\frac{{u – 3}}{{u + 3}}} \right| }+{ C }}
    = {{2\sqrt {x + 9} }+{ 3\ln \left| {\frac{{\sqrt {x + 9} – 3}}{{\sqrt {x + 9} + 3}}} \right| }+{ C.}}
    \]

    Example 2.

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

    Solution.

    We make the following substitution:

    \[
    {\sqrt x = u,\;\;}\Rightarrow
    {x = {u^2},\;\;\;}\;\kern-0.3pt{dx = 2udu.}
    \]

    Then the integral (we denote it by \(I\text{)}\) becomes

    \[
    {I = \int {\frac{{\sqrt x – 1}}{{\sqrt x + 1}}dx} }
    = {\int {\frac{{u – 1}}{{u + 1}}2udu} }
    = {2\int {\frac{{{u^2} – u}}{{u + 1}}du} .}
    \]

    Divide the fraction:

    \[{\frac{{{u^2} – u}}{{u + 1}} }={ u – 2 }+{ \frac{2}{{u + 1}}.}\]

    As a result, we have

    \[
    {I }={ 2\int {\left( {u – 2 + \frac{2}{{u + 1}}} \right)du} }
    = {{2\int {udu} }-{ 4\int {du} }+{ 4\int {\frac{{du}}{{u + 1}}} }}
    = {{\frac{{2{u^2}}}{2} – 4u }+{ 4\ln \left| {u + 1} \right| }+{ C }}
    = {{x – 4\sqrt x }+{ 4\ln \left| {\sqrt x + 1} \right| }+{ C.}}
    \]

    Page 1
    Problems 1-2
    Page 2
    Problems 3-7