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I serious have no idea how should i proceed about doing this question. I've tried IBP or substitution but I can't seems to solve it. Can I have some hints of how should I get started? $\int {1/(x^7 -x)} dx$ These are some of the stuffs i've tried... $IBP: u = \frac {1}{x^6-1}, du = \frac {-5x^6}{x^6-1}, dv = \frac 1x dx, v = ln|x|$ Tried substitution method, but not successful.. |
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The cheater's method is to "observe" that $$ \frac{d}{dx} \ln\left(1-\frac{1}{x^6}\right) = \frac{6}{x^7-x} $$ |
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Partial fractions: $x^7-x=x(x^6-1)=x(x^3-1)(x^3+1)$, and then use the standard factorizations of $a^3-b^3$ and $a^3+b^3$ to split each of the cubic factors into a linear factor and a quadratic factor. |
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One way to look at the problem is to say that it would be easy if the integrand were $$\frac{7x^6-1}{x^7-x},$$and also easy if it were $$\frac{x^6-1}{x^7-x}.$$Now take a linear combination of these to knock out the $x^6$ term in the numerator. |
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As in this example it is useful to extract something of the form $$u=x^a\pm\frac{1}{x^a}$$ $$\int\frac{dx}{x^{7}-x}=\int\frac{1}{x^{4}}\frac{dx}{x^{3}-\frac{1}{x^{3}}}=\int\frac{\left(x^{2}+\frac{1}{x^{4}}\right)dx}{x^{3}-\frac{1}{x^{3}}}-\int\frac{x^{2}dx}{x^{3}-\frac{1}{x^{3}}}=I_1-I_2$$ Now in $I_1$ we may let $$u=x^{3}-\frac{1}{x^{3}}$$ $$du=3\left(x^{2}+\frac{1}{x^{4}}\right)dx$$ So that $$I_1=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}\ln|u|=\frac{1}{6}\ln u^2=\frac{1}{6}\ln\frac{(x^6-1)^2}{x^6}$$ In $I_2$ apparently $v=x^3$ $$I_2=\frac{1}{3}\int\frac{vdv}{v^{2}-1}=\frac{1}{6}\ln\left(v^{2}-1\right)=\frac{1}{6}\ln|x^6-1|$$ Finally, $$I=I_1-I_2=\frac{1}{6}\ln\frac{x^6-1}{x^6}=\frac{1}{6}\ln\left(1-\frac{1}{x^6}\right)$$ |
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