# Integration of Irrational Functions

## Solved Problems

### Example 9.

Evaluate the integral $\int {\frac{{dx}}{{x + \sqrt[3]{x}}}}.$

Solution.

We can write the integral as

$I = \int {\frac{{dx}}{{x + \sqrt[3]{x}}}} = \int {\frac{{dx}}{{x + {x^{\frac{1}{3}}}}}} .$

Since the least common multiple ($$\text{LCM}$$) of the denominators of the fractional powers is equal to $$n = \text{LCM}\left({1,3}\right) = 3,$$ we make the substitution:

$u = {x^{\frac{1}{3}}},\;\; \Rightarrow x = {u^3},\;\;\; dx = 3{u^2}du.$

Thus

$I = \int {\frac{{dx}}{{x + {x^{\frac{1}{3}}}}}} = \int {\frac{{3{u^2}du}}{{{u^3} + {{\left( {{u^3}} \right)}^{\frac{1}{3}}}}}} = 3\int {\frac{{{u^2}du}}{{{u^3} + u}}} = 3\int {\frac{{udu}}{{{u^2} + 1}}}.$

Make the new substitution:

$t = {u^2} + 1,\;\;dt = 2udu,\;\; \Rightarrow udu = \frac{{dt}}{2}.$

$I = 3\int {\frac{{udu}}{{{u^2} + 1}}} = 3\int {\frac{{\frac{{dt}}{2}}}{t}} = 3 \cdot \frac{1}{2}\int {\frac{{dt}}{t}} = \frac{3}{2}\ln \left| t \right| + C = \frac{3}{2}\ln \left( {{u^2} + 1} \right) + C = \frac{3}{2}\ln \left( {{{\left( {{x^{\frac{1}{3}}}} \right)}^2} + 1} \right) + C = \frac{3}{2}\ln \left( {{x^{\frac{2}{3}}} + 1} \right) + C = \frac{3}{2}\ln \left( {\sqrt[3]{{{x^2}}} + 1} \right) + C.$

### Example 10.

Evaluate the integral $\int {\frac{{dx}}{{\sqrt x + \sqrt[3]{x}}}}.$

Solution.

We rewrite the integral in terms of fractional powers of $$x:$$

$I = \int {\frac{{dx}}{{\sqrt x + \sqrt[3]{x}}}} = \int {\frac{{dx}}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}} .$

Then we make the substitution $$u = {x^{\frac{1}{n}}},$$ where $$n$$ is the least common multiple of the denominators:

$n = \text{LCM}\left( {2,3} \right) = 6.$

Hence

$u = {x^{\frac{1}{6}}},\;\; \Rightarrow x = {u^6},\;\; dx = 6{u^5}du,$

so we obtain

$I = \int {\frac{{6{u^5}du}}{{{{\left( {{u^6}} \right)}^{\frac{1}{2}}} + {{\left( {{u^6}} \right)}^{\frac{1}{3}}}}}} = 6\int {\frac{{{u^5}du}}{{{u^3} + {u^2}}}} = 6\int {\frac{{{u^5}du}}{{{u^2}\left( {u + 1} \right)}}} = 6\int {\frac{{{u^3}du}}{{u + 1}}} .$

Since the integrand is an improper function, we use long division:

$\frac{{{u^3}}}{{u + 1}} = {u^2} - u + 1 - \frac{1}{{u + 1}}.$

Now we can easily integrate the last expression:

$I = 6\int {\frac{{{u^3}du}}{{u + 1}}} = 6\int {\left( {{u^2} - u + 1 - \frac{1}{{u + 1}}} \right)dx} = 6\left( {\frac{{{u^3}}}{3} - \frac{{{u^2}}}{2} + u - \ln \left| {u + 1} \right|} \right) + C = 2{u^3} - 3{u^2} + 6u - 6\ln \left| {u + 1} \right| + C.$

Replacing $$u$$ by $${\sqrt[6]{x}}$$ gives:

$I = 2\sqrt x - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln \left| {\sqrt[6]{x} + 1} \right| + C.$

### Example 11.

Evaluate the integral $\int {\frac{{dx}}{{\sqrt[3]{x} + 1}}}.$

Solution.

We make the substitution

$u = {x^{\frac{1}{3}}} = \sqrt[3]{x},\;\; \Rightarrow x = {u^3},\;\; dx = 3{u^2}du.$

Then the integral in $$u-$$terms becomes

$I = \int {\frac{{dx}}{{\sqrt[3]{x} + 1}}} = \int {\frac{{3{u^2}du}}{{u + 1}}} = 3\int {\frac{{{u^2}}}{{u + 1}}du} .$

As the fraction is improper, we perform long division:

$\frac{{{u^2}}}{{u + 1}} = u - 1 + \frac{1}{{u + 1}},$

so we have

$I = 3\int {\left( {u - 1 + \frac{1}{{u + 1}}} \right)du} = \frac{{3{u^2}}}{2} - 3u + 3\ln \left| {u + 1} \right| + C = \frac{{3\sqrt[3]{{{x^2}}}}}{2} - 3\sqrt[3]{x} + 3\ln \left| {\sqrt[3]{x} + 1} \right| + C.$

### Example 12.

Evaluate the integral $\int {\frac{{dx}}{{\sqrt[5]{x} - 1}}}.$

Solution.

We can write the integral as

$\int {\frac{{dx}}{{\sqrt[5]{x} - 1}}} = \int {\frac{{dx}}{{{x^{\frac{1}{5}}} - 1}}} .$

Make the substitution:

$x^{\frac{1}{5}} = u,\;\; \Rightarrow x = {u^5},\;\; dx = 5{u^4}du.$

The integral becomes

$I = \int {\frac{{dx}}{{{x^{\frac{1}{5}}} - 1}}} = \int {\frac{{5{u^4}du}}{{u - 1}}} = 5\int {\frac{{{u^4}du}}{{u - 1}}} .$

Since the degree of the numerator is greater than the degree of the denominator, we divide the numerator by the denominator:

$\frac{{{u^4}}}{{u - 1}} = {u^3} + {u^2} + u + 1 + \frac{1}{{u - 1}}.$

The integral is equal to

$I = 5\int {\left( {{u^3} + {u^2} + u + 1 + \frac{1}{{u - 1}}} \right)du} = 5 \left( {\frac{{{u^4}}}{4} + \frac{{{u^3}}}{3} + \frac{{{u^2}}}{2} + u + \ln \left| {u - 1} \right|} \right) + C = 5 \left( {\frac{{{\sqrt[5]{x^4}}}}{4} + \frac{{{\sqrt[5]{x^3}}}}{3} + \frac{{{\sqrt[5]{x^2}}}}{2} + \sqrt[5]{x} + \ln \left| {\sqrt[5]{x} - 1} \right|} \right) + C.$

### Example 13.

Find the integral $\int {\frac{{dx}}{{\sqrt[3]{x} - \sqrt[4]{x}}}}.$

Solution.

We write the integral as

$I = \int {\frac{{dx}}{{\sqrt[3]{x} - \sqrt[4]{x}}}} = \int {\frac{{dx}}{{{x^{\frac{1}{3}}} - {x^{\frac{1}{4}}}}}} .$

As you can see, the least common multiple $$\left(\text{LCM}\right)$$ of the denominators of the fractional powers is equal to $$n =$$ $$\text{LCM}\left({3,4}\right)$$ $$= 12.$$ Therefore, we make the substitution:

$u = {x^{\frac{1}{{12}}}},\;\; \Rightarrow x = {u^{12}},\;\; dx = 12{u^{11}}du.$

This yields

$I = \int {\frac{{12{u^{11}}du}}{{{{\left( {{u^{12}}} \right)}^{\frac{1}{3}}} - {{\left( {{u^{12}}} \right)}^{\frac{1}{4}}}}}} = \int {\frac{{12{u^{11}}du}}{{{u^4} - {u^3}}}} = 12\int {\frac{{{u^8}du}}{{u - 1}}} .$

The degree of the numerator is greater than the degree of the denominator, therefore, we divide the fraction:

$\frac{{{u^8}}}{{u - 1}} = {u^7} + {u^6} + {u^5} + {u^4} + {u^3} + {u^2} + u + 1 + \frac{1}{{u - 1}}.$

After simple transformations we obtain the final answer:

$I = 12\int {\left( {{u^7} + {u^6} + {u^5} + {u^4} + {u^3} + {u^2} + u + 1 + \frac{1}{{u - 1}}} \right) du} = 12\left( {\frac{{{u^8}}}{8} + \frac{{{u^7}}}{7} + \frac{{{u^6}}}{6} + \frac{{{u^5}}}{5} + \frac{{{u^4}}}{4} + \frac{{{u^3}}}{3} + \frac{{{u^2}}}{2} + u + \ln \left| {u - 1} \right|} \right) + C = \frac{3}{2}{u^8} + \frac{{12}}{7}{u^7} + 2{u^6} + \frac{{12}}{5}{u^5} + 3{u^4} + 4{u^3} + 6{u^2} + 12u + 12\ln \left| {u - 1} \right| + C = \frac{3}{2}{\left( {{x^{\frac{1}{{12}}}}} \right)^8} + \frac{{12}}{7}{\left( {{x^{\frac{1}{{12}}}}} \right)^7} + 2{\left( {{x^{\frac{1}{{12}}}}} \right)^6} + \frac{{12}}{5}{\left( {{x^{\frac{1}{{12}}}}} \right)^5} + 3{\left( {{x^{\frac{1}{{12}}}}} \right)^4} + 4{\left( {{x^{\frac{1}{{12}}}}} \right)^3} + 6{\left( {{x^{\frac{1}{{12}}}}} \right)^2} + 12{x^{\frac{1}{{12}}}} + 12\ln \left| {{x^{\frac{1}{{12}}}} - 1} \right| + C = \frac{3}{2}{x^{\frac{3}{2}}} + \frac{{12}}{7}{x^{\frac{7}{{12}}}} + 2{x^{\frac{1}{2}}} + \frac{{12}}{5}{x^{\frac{5}{{12}}}} + 3{x^{\frac{1}{3}}} + 4{x^{\frac{1}{4}}} + 6{x^{\frac{1}{6}}} + 12{x^{\frac{1}{{12}}}} + 12\ln \left| {{x^{\frac{1}{{12}}}} - 1} \right| + C = \frac{3}{2}\sqrt[3]{{{x^2}}} + \frac{{12}}{7}\sqrt[{12}]{{{x^7}}} + 2\sqrt x + \frac{{12}}{5}\sqrt[{12}]{{{x^5}}} + 3\sqrt[3]{x} + 4\sqrt[4]{x} + 6\sqrt[6]{x} + 12\sqrt[{12}]{x} + 12\ln \left| {\sqrt[{12}]{x} - 1} \right| + C.$

### Example 14.

Evaluate the integral $\int {\frac{{dx}}{{\sqrt[3]{{{x^2}}} - \sqrt x }}}.$

Solution.

We write the integral in terms of fractional powers of $$x:$$

$I = \int {\frac{{dx}}{{\sqrt[3]{{{x^2}}} - \sqrt x }}} = \int {\frac{{dx}}{{{x^{\frac{2}{3}}} - {x^{\frac{1}{2}}}}}} .$

As the least common multiple of the denominators is

$n = LCM\left( {3,2} \right) = 6,$

the substitution is pretty clear:

$u = {x^{\frac{1}{6}}} = \sqrt[6]{x},\;\; \Rightarrow x = {u^6},\;\; dx = 6{u^5}du.$

Then the integral in $$u-$$terms becomes

$I = \int {\frac{{6{u^5}du}}{{{{\left( {{u^6}} \right)}^{\frac{2}{3}}} - {{\left( {{u^6}} \right)}^{\frac{1}{2}}}}}} = 6\int {\frac{{{u^5}du}}{{{u^4} - {u^3}}}} = 6\int {\frac{{{u^5}du}}{{{u^3}\left( {u - 1} \right)}}} = 6\int {\frac{{{u^2}du}}{{u - 1}}} .$

The integrand is an improper rational fraction, so we perform long division before integrating:

$\frac{{{u^2}}}{{u - 1}} = u + 1 + \frac{1}{{u - 1}}.$

Hence

$I = 6\int {\left( {u + 1 + \frac{1}{{u - 1}}} \right)du} = 6\left[ {\frac{{{u^2}}}{2} + u + \ln \left| {u - 1} \right|} \right] + C = 3{u^2} + 6u + 6\ln \left| {u - 1} \right| + C = 3\sqrt[3]{x} + 6\sqrt[6]{x} + 6\ln \left| {\sqrt[6]{x} - 1} \right| + C.$

### Example 15.

Evaluate $\int {\frac{{dx}}{{\sqrt {\sqrt x - 2} }}}.$

Solution.

Make the substitution:

$\sqrt x - 2 = {u^2},\;\; \Rightarrow \sqrt x = {u^2} + 2,\;\; \Rightarrow x = {\left( {{u^2} + 2} \right)^2} = {u^4} + 4{u^2} + 4,\;\; \Rightarrow dx = \left( {4{u^3} + 8u} \right)du.$

This yields

$I = \int {\frac{{dx}}{{\sqrt {\sqrt x - 2} }}} = \int {\frac{{\left( {4{u^3} + 8u} \right)du}}{u}} = 4\int {\left( {{u^2} + 2} \right)du} = \frac{{4{u^3}}}{3} + 8u + C = \frac{4}{3}\sqrt {{{\left( {\sqrt x - 2} \right)}^3}} + 8\sqrt {\sqrt x - 2} + C.$

### Example 16.

Calculate the integral $\int {\sqrt {{e^x} + 1}\,dx}.$

Solution.

We make the substitution

${e^x} + 1 = {u^2},\;\; \Rightarrow {e^x}dx = 2udu,\;\; \Rightarrow dx = \frac{{2udu}}{{{e^x}}} = \frac{{2udu}}{{{u^2} - 1}}.$

As a result, we have

$I = \int {\sqrt {{e^x} + 1} dx} = \int {u\frac{{2udu}}{{{u^2} - 1}}} = 2\int {\frac{{{u^2}du}}{{{u^2} - 1}}} = 2\int {\frac{{{u^2} - 1 + 1}}{{{u^2} - 1}}du} = 2\int {\left( {1 + \frac{1}{{{u^2} - 1}}} \right)du} = 2\int {du} - 2\int {\frac{{du}}{{1 - {u^2}}}} = 2u - 2 \cdot \frac{1}{2}\ln \left| {\frac{{1 + u}}{{1 - u}}} \right| + C = 2u - \ln \left| {\frac{{1 + u}}{{1 - u}}} \right| + C = 2\sqrt {{e^x} + 1} - \ln \left| {\frac{{1 + \sqrt {{e^x} + 1} }}{{1 - \sqrt {{e^x} + 1} }}} \right| + C.$

### Example 17.

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

Solution.

Making the substitution

$t = \sqrt x ,\;\; \Rightarrow x = {t^2},\;\;dx = 2tdt,$

we can write the integral in the form:

$I = \int {{e^{\sqrt x }}dx} = \int {{e^t} \cdot 2tdt} = 2\int {t{e^t}dt} .$

Now we use integration by parts to get:

$I = 2\int {t{e^t}dt} = \left[ {\begin{array}{*{20}{l}} {u = t}\\ {u^\prime = 1}\\ {v^\prime = {e^t}}\\ {v = e} \end{array}} \right] = 2\left( {t{e^t} - \int {{e^t}dt} } \right) = 2\left( {t{e^t} - {e^t}} \right) + C = 2\left( {t - 1} \right){e^t} + C = 2\left( {\sqrt x - 1} \right){e^{\sqrt x }} + C.$