Calculus

Integration of Functions

Integration of Functions Logo

Integration of Irrational Functions

  • Certain types of integrals containing irrational expressions can be reduced to integrals of rational functions by making an appropriate substitution.

    Such transformations of an integral is called its rationalization.

    In this topic, we look at a few useful substitutions that can help with irrational integrals.

    Case \(1.\) Integrals Involving Fractional Powers of \(x\)

    To integrate a function that contains only one irrational expression of the form \({x^{\large{\frac{m}{n}}\normalsize}},\) we make the substitution \(u = {x^{\large{\frac{1}{n}}\normalsize}}.\)

    If an irrational function contains more than one rational power of \(x,\) we use the substitution \(u = {x^{\large{\frac{1}{n}}\normalsize}},\) where \(n\) is the least common multiple (LCM) of the denominators of all fractional powers of x.

    Case \(2.\) Integrals Involving \(\sqrt[n]{{\large{\frac{{ax + b}}{{cx + d}}}\normalsize}}\)

    An expression of the form \(\sqrt[n]{{\large{\frac{{ax + b}}{{cx + d}}}\normalsize}}\) can be integrated using the substitution \(u = {\left( {\large{\frac{{ax + b}}{{cx + d}}}\normalsize} \right)^{\large{\frac{1}{n}}\normalsize}},\) where \(a, b, c, d\) are real numbers.

    These substitutions reduce the integrals to rational functions in \(u.\)

    Note that integrals containing radicals of the form \(\sqrt {{a^2} – {x^2}},\) \(\sqrt {{a^2} + {x^2}},\) and \(\sqrt {{x^2} – {a^2}}\) can be evaluated with the help of trigonometric and hyperbolic substitutions.


  • 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

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

    Example 3

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

    Example 4

    Evaluate the integral \(\int {\sqrt[3]{{5x – 1}}dx}.\)

    Example 5

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

    Example 6

    Find the integral \(\int {x\sqrt {2x – 3} dx}.\)

    Example 7

    Evaluate the integral \(\int {\large{\frac{{dx}}{{x\sqrt {x – 4} }}}\normalsize}.\)

    Example 8

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

    Example 9

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

    Example 10

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

    Example 11

    Evaluate the integral \(\int {\large{\frac{{dx}}{{\sqrt[3]{x} + 1}}}\normalsize}.\)

    Example 12

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

    Example 13

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

    Example 14

    Evaluate the integral \(\int {\large{\frac{{dx}}{{\sqrt[3]{{{x^2}}} – \sqrt x }}}\normalsize}.\)

    Example 15

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

    Example 16

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

    Example 17

    Evaluate the integral \(\int {{e^{\sqrt x }}dx}.\)

    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.

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

    Solution.

    Using the substitution

    \[{u = {x^{\frac{1}{2}}} = \sqrt x,}\;\; \Rightarrow {x = {u^2},\;\;}\kern0pt{dx = 2udu,}\]

    we obtain

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

    Example 3.

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

    Solution.

    We substitute

    \[{u = \sqrt x ,}\;\; \Rightarrow {x = {u^2},\;\;}\kern0pt{dx = 2udu.}\]

    Hence

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

    Example 4.

    Evaluate the integral \(\int {\sqrt[3]{{5x – 1}}dx}.\)

    Solution.

    We use the substitution:

    \[{u = {\left( {5x – 1} \right)^{\frac{1}{3}}} }={ \sqrt[3]{{5x – 1}},}\;\; \Rightarrow {5x – 1 = {u^3},}\;\; \Rightarrow {5x = {u^3} + 1,}\Rightarrow {x = \frac{{{u^3} + 1}}{5},\;\;}\kern0pt{dx = \frac{{3{u^2}du}}{5}.}\]

    Then

    \[{\int {\sqrt[3]{{5x – 1}}dx} }={ \int {u \cdot \frac{{3{u^2}du}}{5}} }={ \frac{3}{5}\int {{u^3}du} }={ \frac{3}{5} \cdot \frac{{{u^4}}}{4} + C }={ \frac{{3{u^4}}}{{20}} + C }={ \frac{{3\sqrt[3]{{{{\left( {5x – 1} \right)}^4}}}}}{{20}} + C.}\]

    Example 5.

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

    Solution.

    We make the substitution

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

    Hence

    \[{x + 1 = {u^2},}\;\; \Rightarrow {x = {u^2} – 1,\;\;}\kern0pt{dx = 2udu.}\]

    The integral in \(u-\)terms becomes

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

    Example 6.

    Find the integral \(\int {x\sqrt {2x – 3} dx}.\)

    Solution.

    Using the substitution

    \[u = {\left( {2x – 3} \right)^{\frac{1}{2}}} = \sqrt {2x – 3},\]

    we have

    \[{2x + 3 = {u^2},}\;\; \Rightarrow {2x = {u^2} + 3,}\;\; \Rightarrow {x = \frac{{{u^2} + 3}}{2},\;\;}\kern0pt{dx = udu.}\]

    Now we can easily calculate the integral:

    \[{\int {x\sqrt {2x – 3} dx} }={ \int {\frac{{{u^2} + 3}}{2} \cdot u \cdot udu} }={ \frac{1}{2}\int {\left( {{u^4} + 3{u^2}} \right)du} }={ \frac{1}{2}\left[ {\frac{{{u^5}}}{5} + 3 \cdot \frac{{{u^3}}}{3}} \right] + C }={ \frac{{{u^5}}}{{10}} + \frac{{{u^3}}}{2} + C }={ \frac{{\sqrt {{{\left( {2x – 3} \right)}^5}} }}{{10}} }+{ \frac{{\sqrt {{{\left( {2x – 3} \right)}^3}} }}{2} }+{ C.}\]

    Page 1
    Problems 1-6
    Page 2
    Problems 7-17