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Rationalization of the Integrand

April 5th, 2009 |

Author: Merrin

When faced with an integrand where there are terms containing roots of x or roots of complex expressions, it may be helpful to perform a substitution which converts the integrand into a rational function. First identify the troublesome term, and then try a u substitution of specifically that term.

Example 1. Here is a typical example where rationalization of the integrand by u substitution is possible.

$\displaystyle \int \cfrac{dx}{a+\sqrt{x}}$

Solution 1. The integrand contains a denominator we don’t know how to integrate in this form. We attempt to rationalize the integrand by making a u substitution.

$u=\sqrt{x}$

We must first calculate du in terms of dx and remove all traces of x.

$du= \cfrac{1}{2x^{1/2}}dx=\cfrac{1}{2u} dx$
$dx = 2udu$

Once the integral is rationalized and transformed in terms of u, we can try to take the integral using our standard methods.

$\displaystyle \int \cfrac{2udu}{a+u} = 2\displaystyle \int \cfrac{(a+u-a)du}{a+u} =2 \displaystyle \int du-2a\displaystyle \int \cfrac{du}{a+u}$
$=2u-2a\ln \left|a+u\right|+C=2\sqrt{x}-2a\ln\left|a+\sqrt{x}\right|+C$

We have used a trick where we add and subtract the same thing in the numerator to get a linear term and a logarithmic term. This trick is worth remembering. In the end, we substitute back the square root of x for u and we have solved the integral.

$\displaystyle \int \cfrac{dx}{a+\sqrt{x}}=2\sqrt{x} -2a\ln \left|a+\sqrt{x} \right|+C$

Let us try an alternate substitution and see what happens.

$\displaystyle \int \cfrac{dx}{a+\sqrt{x} } \qquad u=a+\sqrt{x} \qquad du=\cfrac{dx}{2\sqrt{x} } \qquad dx=2(u-a)du$
$\int \cfrac{dx}{a+\sqrt{x} } =\displaystyle \int \cfrac{2(u-a)}{u} du =2\displaystyle \int du -2a\displaystyle \int \cfrac{du}{u} =2u-2a\ln \left|u\right|+C$
$\int \cfrac{dx}{a+\sqrt{x} } =2(a+\sqrt{x} )-2a\ln \left|a+\sqrt{x} \right|+C$

The answers are the same up to a constant which can be absorbed into the integration constant. Sometimes several different substitutions can lead to the same final answer. This second substitution doesn’t require knowing the add and subtract trick that allowed us to solve the first problem, so perhaps it is better.

Example 2. Try this integral where the integrand would be rational except for a nasty term.

$\displaystyle \int x^{3} \sqrt[{4}]{x^{2} +a^{2}} dx$

Solution 2. We perform the suggested substitution.

$u=\sqrt[{4}]{x^{2} +a^{2} }$

We have a bit of calculation to transform the integral.

${du} = {\cfrac{1}{4} \cfrac{2x}{(x^{2} +a^{2} )^{3/4} } dx\quad \qquad dx=\cfrac{2(x^{2} +a^{2} )^{3/4} }{x} du}$

${\displaystyle \int x^{3} \sqrt[{4}]{x^{2} +a^{2} } dx} = {2\displaystyle \int x^{3} \cfrac{(x^{2} +a^{2} )^{3/4} }{x} du=2\displaystyle \int (u^4 -a^2)u^3}du$

This integral is straightforward to calculate since it there is just a polynomial in the integrand. After we have integrated, we substitute our formula for u in terms of x.

${2\displaystyle \int (u^{7} -a^{2} u^{3} )du } = {\cfrac{1}{4} u^{8} -\cfrac{1}{2} a^{2} u^{4} +C}$
$\int x^{3} \sqrt[{4}]{x^{2} +a^{2} } dx} = {\cfrac{1}{4} (x^{2} +a^{2} )^{2} -\cfrac{1}{2} (x^{2} +a^{2} )+C$

Example 3. For the more spectacular integrals try to substitute the least common multiple of similarly looking radicals.

$\displaystyle \int \cfrac{dx}{x^{1/5} +x^{1/2} }$

Solution 3. The least common multiple is

$u=x^{1/10}$

$\int \cfrac{dx}{x^{1/5} +x^{1/2} } =\displaystyle \int \cfrac{10u^{9} du}{u^{2} +u^{5} } =\displaystyle \int \cfrac{10u^{7} }{1+u^{3} } du$
${u^{3} +1=(u+1)(u^{2} -u+1)}$