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$$\int\frac{dx}{x-\sqrt[4]{x}}$$

given the substitution $x=u^{4}, dx=4u^{3}du$

$$=\int\frac{4u^{3}du}{u^{4}-u}=\int\frac{4u^{3}du}{u(u^{3}-1)}=\int\frac{4u^{2}du}{(u^{3}-1)}$$

At this point I believe I can factor the denominator and use partial fractions

$$\int\frac{4u^{2}du}{(u-1)(u^{2}+u+1)}=\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du$$

by substituting in $u=1$ and ignoring the factor in the denominator $(u-1)$ (Grubby Thumb Rule) $A=\frac{4*1^{2}}{1^{2}+1+1}, A=\frac{4}{3}$

and substituting arbitrarily $u=0$ into the expression $$4u^{2}=\frac{4}{3}(u^2+u+1)+B(u-1)$$ $$0=\frac{4}{3}(1)-B, B=\frac{4}{3}$$

therefore

$$\int\bigg(\frac{A}{u-1}+\frac{B}{u^2+u+1}\bigg)du=\int\bigg(\frac{4}{3(u-1)}+\frac{4}{3(u^2+u+1)}\bigg)du$$

I am able to solve this however I dont think the value for B is correct

I have the following questions:

1: Am I able to use partial fractions even though the denominator of B is $u^{2}+u+1$

2: Did I do the substitution properly?

3: am I able to do what I did to find the value for A?

Thanks Stax

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  • $\begingroup$ Are you told that you have to do it with partial fractions? Because it's much easier without. $\endgroup$
    – David
    Feb 5, 2016 at 1:00
  • $\begingroup$ yes, I am supposed to use the provided substitution and then solve by partial fractions. I would be interested in the solution as you describe it though $\endgroup$
    – helpmeh
    Feb 5, 2016 at 1:01
  • $\begingroup$ I wonder about the wisdom of telling people to use partial fractions for a problem that can be done more easily by simpler methods. I posted such a method below. $\qquad$ $\endgroup$ Feb 5, 2016 at 1:09

2 Answers 2

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If you have to use partial fractions, you need to start with $$\frac{4u^{2}}{(u-1)(u^{2}+u+1)}=\frac{A}{u-1}+\frac{Bu+C}{u^2+u+1}$$ and then follow pretty much the procedure you have tried.

The easy way to evaluate $$\int\frac{4u^{2}du}{u^{3}-1}=\frac43\int\frac{3u^{2}du}{u^{3}-1}$$ is by substituting $v=u^3-1$, which gives the answer immediately.

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  • $\begingroup$ Thank you! I might in fact use the easy way out of spite ;) $\endgroup$
    – helpmeh
    Feb 5, 2016 at 1:10
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$$ \int\frac{4u^{2}du}{u^3-1} = \frac 4 3 \int \frac {dw} w, \text{ etc.} $$

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