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mllamontagne
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Homework Statement
[itex]\int\frac{1}{x\sqrt[3]{x+1}}[/itex]dx (That's a cubic root in the denominator, by the way. Not an x cubed.)
Hello mllamontagne. Welcome to PF !mllamontagne said:Homework Statement
[itex]\int\frac{1}{x\sqrt[3]{x+1}}[/itex]dx (That's a cubic root in the denominator, by the way. Not an x cubed.)
The Attempt at a Solution
I thought possibly partial fractions, but I've never seen it done with a root in the denominator. Integration by parts was unsuccessful. Thanks so much if anyone can show me the steps for this. Must be able to broken up into several bits that integrate to logs or trig functions or something, but I can't seem to get it.
mllamontagne said:As you suggest, Partial fractions can be used to break up
3∫[itex]\frac{u}{(u-1)(u^{2}+u+1)}[/itex]du with u=[itex]\sqrt[3]{x+1}[/itex]
to ∫[itex]\frac{1}{u-1}[/itex]du +∫[itex]\frac{1}{u^{2}+u+1}[/itex] du -∫[itex]\frac{u}{u^{2}+u+1}[/itex] du
the first integral is a simple natural log, the second is and arctan, but now the third one gives me trouble. how would I accomplish the operation below?
∫[itex]\frac{u}{u^{2}+u+1}[/itex] du
The purpose of using partial fractions for a cubic root in the denominator of an integrand is to simplify the integration process. This allows for easier evaluation of the integral and can also help in finding the antiderivative.
The process of decomposing a cubic root in the denominator of an integrand into partial fractions involves finding the roots of the denominator polynomial, writing the partial fraction in terms of the roots, and equating the coefficients of like powers of the variable in the numerator and denominator.
Yes, partial fractions can be used to integrate any type of cubic root in the denominator of an integrand. However, the process may vary slightly depending on the type of cubic root (e.g. simple, repeated, or complex roots).
One limitation of using partial fractions for a cubic root in the denominator of an integrand is that it can only be applied to rational functions. Additionally, the degree of the numerator must be less than the degree of the denominator for the decomposition to be possible.
Yes, partial fractions for a cubic root in the denominator of an integrand can be used in real-life applications, particularly in engineering and physics problems that require the evaluation of integrals. It can also be used in the process of solving differential equations.