Select Page

# 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