Rational Integration

**juantheron** · 10 Sep 2018, 6:07 PM

Evaluation of $\displaystyle \int\frac{x^5+3}{(x-5)^3\cdot (x-1)}dx$

Although i have solved it using partial fraction. But it is very lengthy

so please explain me any bettter method. Thanks.

**aa180** · 13 Sep 2018, 12:01 AM

Originally Posted by juantheron

Evaluation of $\displaystyle \int\frac{x^5+3}{(x-5)^3\cdot (x-1)}dx$

Although i have solved it using partial fraction. But it is very lengthy

so please explain me any bettter method. Thanks.

How long exactly is your partial fraction approach? I do have an alternative method but it may be even longer than the way you did it.

**HeroWise** · 13 Sep 2018, 12:03 AM

Im curious what is this other method u are talking of

**aa180** · 13 Sep 2018, 12:06 AM

Originally Posted by HeroWise

Im curious what is this other method u are talking of

It will take quite some time for me to type it all up so not sure if I'll do it tonight. Though I should mention that my method still uses partial fractions at one point, but it's a much simpler case that can be done by inspection.

**aa180** · 13 Sep 2018, 7:01 PM

$\begin{align*} \int\frac{x^{5}+3}{(x-5)^{3}(x-1)}dx &= \int\frac{x^{5}-1+4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{x^{5}-1}{(x-5)^{3}(x-1)}dx + \int\frac{4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{(x-1)(x^{4}+x^{3}+x^{2}+x+1)}{(x-5)^{3}(x-1)}dx + \int\frac{4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{x^{4}+x^{3}+x^{2}+x+1}{(x-5)^{3}}dx + \int\frac{4}{(x-5)^{3}(x-1)}dx \\ \end{align*}$

$\text{Let } I_{1} = \int\frac{x^{4}+x^{3}+x^{2}+x+1}{(x-5)^{3}}dx, \text{ and } I_{2} = \int\frac{4}{(x-5)^{3}(x-1)}dx. \text{ Consider } I_{1}:$

$\text{Set } u = x^{4}+x^{3}+x^{2}+x+1 \text{ and } v' = \frac{1}{(x-5)^{3}} = (x-5)^{-3} \text{ and integrate by parts.} \\ \text{Once you have } (x-5)^{1} \text{ in the denominator (after second application of IBP), simply perform a} \\ \text{polynomial division rather than integrate by parts a third time. Integrating the result and} \\ \text{your final answer should be:}$

$I_{1} = -\frac{(x^{4}+x^{3}+x^{2}+x+1)}{2(x-5)^{2}} - \frac{4x^{3}+3x^{2}+2x+1}{2(x-5)} + 3x^{2} + 33x + 166\ln|x-5| + C_{1}$

$\text{Now consider } I_{2}:$

$\begin{align*} I_{2} &= \int\frac{4}{(x-5)^{3}(x-1)}dx \\ &= \int\frac{4}{[(x-3)-2)]^{3}[(x-3)+2]}dx \\ \end{align*}$

$\text{Let } u = x-3 \Longrightarrow dx = du. \text{ Hence:}$

$\begin{align*} I_{2} &= \int\frac{4}{(u-2)^{3}(u+2)}du \\ &= \int\frac{4}{(u-2)(u+2)}\cdot\frac{1}{(u-2)^{2}}du \\ &= \int(\frac{1}{u-2}-\frac{1}{u+2})\cdot\frac{1}{(u-2)^{2}}du \\ &= \int\frac{1}{(u-2)^{3}}du - \int\frac{1}{(u+2)(u-2)^{2}}du \\ &= -\frac{1}{2(u-2)^{2}} - \frac{1}{4}\int\frac{4}{(u+2)(u-2)}\cdot\frac{1}{u-2}du \\ &= -\frac{1}{2(u-2)^{2}} - \frac{1}{4}\int(\frac{1}{u-2} - \frac{1}{u+2})\cdot\frac{1}{u-2}du \\ &= -\frac{1}{2(u-2)^{2}} - \frac{1}{4}\int\frac{1}{(u-2)^{2}}du + \frac{1}{4}\int\frac{1}{(u+2)(u-2)}du \\ &= -\frac{1}{2(u-2)^{2}} + \frac{1}{4(u-2)} + \frac{1}{16}\int(\frac{1}{u-2} - \frac{1}{u+2})du \\ &= -\frac{1}{2(u-2)^{2}} + \frac{1}{4(u-2)} + \frac{1}{16}\ln|u-2| - \frac{1}{16}\ln|u+2| + C_{2} \\ &= -\frac{1}{2(x-5)^{2}} + \frac{1}{4(x-5)} + \frac{1}{16}\ln|x-5| - \frac{1}{16}\ln|x-1| + C_{2} \\ \end{align*}$

$\text{Combining } I_{1} \text{ and } I_{2} \text{ gives the correct result (Note that we can simplify things a little} \\ \text{further if we choose to do so).}$