Calculus

Integration of Functions

Integration of Irrational Functions

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

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 on problem description to see 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