$\int\frac{1}{x(x⁷+1)}dx$ without using substitution or by parts. My problem is how to solve the algebra in it which methods are available to solve it? $1=(x^7+1)-x^7$ then we substitute in 1 and the integration became easy. I don't know how this is possible.
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$\begingroup$ Please make sure my edit is correct. You wrote $1/x(x^7+1)$ which could also mean $\frac{x^7+1}{x}$. $\endgroup$– panciniCommented Oct 9, 2016 at 21:10
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$\begingroup$ Please learn MathJax. Formatting tips here. $\endgroup$– Em.Commented Oct 9, 2016 at 21:11
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1$\begingroup$ How exactly do you expect to be able to evaluate this with only algebra? Substitution and integration by parts are the easiest methods to evaluate antiderivatives. Without them, all you're left with is much harder techniques. $\endgroup$– user137731Commented Oct 9, 2016 at 21:11
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1 Answer
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Because I don't think it'll be possible to evaluate your antiderivative with any technique simpler than substitution or integration by parts, here's a hint that'll let you solve by substitution.
With partial fraction decomposition, you can show that $$\require{enclose}\enclose{box}{\frac{1}{x(x^7+1)} = \frac{1}{x} - \frac{x^6}{x^7+1}}$$
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$\begingroup$ Actually now this doesn't need substitutions at all. Now the antiderivative is straight forward by the fact that $\int (f'/f)=\ln f$ $\endgroup$– ExtremalCommented Oct 9, 2016 at 21:53
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$\begingroup$ If you're recognizing the result of the chain rule, you're essentially using substitution, IMO. $\endgroup$– user137731Commented Oct 9, 2016 at 22:57
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$\begingroup$ 1=(x⁷+1)-x⁷ can you explain this? $\endgroup$ Commented Oct 12, 2016 at 21:15
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$\begingroup$ @JoãoNogueira I agree that $1 = (b+1)-b$ for any $b$. What do you mean? $\endgroup$– user137731Commented Oct 12, 2016 at 21:25
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$\begingroup$ I mean if we put this expression in the integral we can solve it then we only need is split it $\endgroup$ Commented Oct 12, 2016 at 21:42