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 m/n, we make the substitution u = x 1/n.
If an irrational function contains more than one rational power of x, we use the substitution u = x 1/n, where n is the least common multiple (LCM) of the denominators of all fractional powers of x.
Case 2. Integrals Involving \(\sqrt[n]{{\frac{{ax + b}}{{cx + d}}}}\)
An expression of the form \(\sqrt[n]{{\frac{{ax + b}}{{cx + d}}}}\) can be integrated using the substitution \(u = {\left( {\frac{{ax + b}}{{cx + d}}} \right)^{\frac{1}{n}}},\) 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
Example 1.
Find the integral \[\int {{\frac{{\sqrt {x + 9} }}{x}} dx}.\]
Solution.
We make the substitution:
\[u = {\left( {x + 9} \right)^{\frac{1}{2}}},\;\; \Rightarrow x + 9 = {u^2},\;\; \Rightarrow x = {u^2} - 9,\;\;\; dx = 2udu.\]
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 \[\int {\frac{{dx}}{{x - \sqrt x }}}.\]
Solution.
Using the substitution
\[u = {x^{\frac{1}{2}}} = \sqrt x,\;\; \Rightarrow x = {u^2},\;\;dx = 2udu,\]
we obtain
\[\int {\frac{{dx}}{{x - \sqrt x }}} = \int {\frac{{2udu}}{{{u^2} - u}}} = 2\int {\frac{{\cancel{u}du}}{{\cancel{u}\left( {u - 1} \right)}}} = 2\int {\frac{{du}}{{u - 1}}} = 2\ln \left| {u - 1} \right| + C = 2\ln \left| {\sqrt x - 1} \right| + C.\]
Example 3.
Evaluate the integral \[\int {\frac{{dx}}{{\sqrt x + 1}}}.\]
Solution.
We substitute
\[u = \sqrt x ,\;\; \Rightarrow x = {u^2},\;\;dx = 2udu.\]
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:
\[u = {\left( {5x - 1} \right)^{\frac{1}{3}}} = \sqrt[3]{{5x - 1}},\;\; \Rightarrow 5x - 1 = {u^3},\;\; \Rightarrow 5x = {u^3} + 1,\;\; \Rightarrow x = \frac{{{u^3} + 1}}{5},\;\; dx = \frac{{3{u^2}du}}{5}.\]
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 \[\int {\frac{x}{{\sqrt {x + 1} }} dx}.\]
Solution.
We make the substitution
\[u = {\left( {x + 1} \right)^{\frac{1}{2}}} = \sqrt {x + 1} .\]
Hence
\[x + 1 = {u^2},\;\; \Rightarrow x = {u^2} - 1,\;\;dx = 2udu.\]
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
\[2x + 3 = {u^2},\;\; \Rightarrow 2x = {u^2} + 3,\;\; \Rightarrow x = \frac{{{u^2} + 3}}{2},\;\;dx = udu.\]
Now we can easily calculate the integral:
\[\int {x\sqrt {2x - 3} dx} = \int {\frac{{{u^2} + 3}}{2} \cdot u \cdot udu} = \frac{1}{2}\int {\left( {{u^4} + 3{u^2}} \right)du} = \frac{1}{2}\left[ {\frac{{{u^5}}}{5} + 3 \cdot \frac{{{u^3}}}{3}} \right] + C = \frac{{{u^5}}}{{10}} + \frac{{{u^3}}}{2} + C = \frac{{\sqrt {{{\left( {2x - 3} \right)}^5}} }}{{10}} + \frac{{\sqrt {{{\left( {2x - 3} \right)}^3}} }}{2} + C.\]
Example 7.
Evaluate the integral \[\int {\frac{{dx}}{{x\sqrt {x - 4} }}}.\]
Solution.
We use the substitution
\[u = {\left( {x - 4} \right)^{\frac{1}{2}}} = \sqrt {x - 4} ,\;\; \Rightarrow x - 4 = {u^2},\;\; \Rightarrow x = 4 + {u^2},\;\; dx = 2udu.\]
This yields:
\[\int {\frac{{dx}}{{x\sqrt {x - 4} }}} = \int {\frac{{2\cancel{u}du}}{{\left( {4 + {u^2}} \right)\cancel{u}}}} = 2\int {\frac{{du}}{{4 + {u^2}}}} = 2\int {\frac{{du}}{{{2^2} + {u^2}}}} = 2 \cdot \frac{1}{2}\arctan \frac{u}{2} + C = \arctan \frac{u}{2} + C = \arctan \frac{{\sqrt {x - 4} }}{2} + C.\]
Example 8.
Calculate the integral \[\int {{\frac{{\sqrt x - 1}}{{\sqrt x + 1}}} dx}.\]
Solution.
We make the following substitution:
\[\sqrt x = u,\;\; \Rightarrow x = {u^2},\;\;\; dx = 2udu.\]
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
\[I = 2\int {\left( {u - 2 + \frac{2}{{u + 1}}} \right)du} = 2\int {udu} - 4\int {du} + 4\int {\frac{{du}}{{u + 1}}} = \frac{{2{u^2}}}{2} - 4u + 4\ln \left| {u + 1} \right| + C = x - 4\sqrt x + 4\ln \left| {\sqrt x + 1} \right| + C.\]
See more problems on Page 2.
