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How do I solve this integral using partial fractions? $$\int\frac{3}{x^3+5}dx$$

My progress is that the numerator would be of the form $Ax^2 + Bx + C$ But then $A$ and $B = 0$ so I am stuck right where I started..

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closed as off-topic by Eevee Trainer, Nosrati, José Carlos Santos, Song, Cesareo Mar 15 at 0:19

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    $\begingroup$ Hi and welcome to MSE! I want to remind you that it is generally preferred you include context when asking a question here (which can include: where this problem came from, your own attempts, and a specific idea as to where you're stuck) - it also lets us help you better! As is, your question is little more than an isolated problem, and thus likely to get a lot of downvotes and closed. Feel free to edit the context into your post though! Here's a useful link: asking a good question $\endgroup$ – Eevee Trainer Mar 14 at 21:43
  • $\begingroup$ You have to factor $x^2+5$ into irreducible factors first. $\endgroup$ – Bernard Mar 14 at 21:44
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    $\begingroup$ Oh, that is sad (when the author restricts to a specific method). $\endgroup$ – Zacky Mar 14 at 21:48
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    $\begingroup$ $x^3+a^3=(x+a)(x^2-ax+a^2)$ so replace $a$ with $5^{^\frac{1}{3}}$ $\endgroup$ – randomgirl Mar 14 at 21:52
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    $\begingroup$ The above substitution of $ t=\frac{1-y}{1+y}$ works nice because the third power gets canceled: $$\frac{1}{t^3+1}\rightarrow \frac{1}{\frac{(1-y)^3 +(1+y)^3}{(1+y)^3}}\frac{2}{(1+y)^2}=\frac{(1+y)^3}{1-3y+3y^2-y^3 +1+3y +3y^2 +y^3}\frac{2}{(1+y)^2}=\frac{1+y}{3y^2+1}$$ Note: I forgot the minus sign, also $y=\frac{\sqrt[3]{5}- x}{\sqrt[3]{5}+ x}$. $\endgroup$ – Zacky Mar 14 at 22:07
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We have

$$\int\dfrac{3}{x^3+5}dx=\int\dfrac{3}{(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{5}^2)}dx.$$

Now, you could try to solve it from here by partial fractions. But you could also find the complex roots of $x^2-\sqrt[3]{5}x+\sqrt[3]{5}^2$. Lets refer to them as $x_1$ and $x_2$.

Then we can decompose the fraction as

$$\dfrac{1}{x^3+5}=\dfrac{\alpha_1}{x+\sqrt[3]{5}}+\dfrac{\alpha_2}{x-x_1}+\dfrac{\alpha_3}{x-x_2}.$$

At this point, you can use Euler's quick method for partial fractions. Multiply by the denominator and set $x$ to the root of the denominator. E.g. for $\alpha_1$

$$\alpha_1=\left.\dfrac{x+\sqrt[3]{5}}{x^3+5}\right|_{x=-\sqrt[3]{5}}=\left.\dfrac{1}{3x^2}\right|_{x=-\sqrt[3]{5}}$$

I used Johann Bernoulli's limit law (falsely called l'Hospital's rule). Integrating the partial fractions expression should be quiet easy.

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  • $\begingroup$ +1 for educating me that my life was a lie $\endgroup$ – Stan Tendijck Mar 14 at 22:11
  • $\begingroup$ Your decomposition does not work, simply because $x^3+5$ does not have 3 distinct real roots. Instead you should have $$ \frac{1}{x^3+5} = \frac{\alpha_1}{x+\sqrt[3]{5}}+\frac{\alpha_2 x + \alpha_3}{x^2-\sqrt[3]{5}x +(\sqrt[3]{5})^2} $$ The problem with the "quick" method is that it will still give you (wrong) results even if you don't setup the decomposition properly. $\endgroup$ – PierreCarre Mar 14 at 22:59
  • $\begingroup$ @PierreCarre What do you mean by wrong? The polynomial has three roots (one real and two complex). $\endgroup$ – MachineLearner Mar 15 at 6:27
  • $\begingroup$ What is the point of integrating complex rational functions in such a simple situation? Overall, the integration process will be more tedious. Regarding the "quick" method, from my experience, students tend to come up with creative versions of it and, contrary to the method of undetermined coefficients, there may be no alert to the fact that you are getting a wrong decomposition. $\endgroup$ – PierreCarre Mar 15 at 7:53
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Hints:

By a rescaling of the variable, we reduce to

$$\int\frac{dx}{x^3+1}=\int\frac{dx}{(x+1)(x^2-x+1)}=\frac13\int\dfrac{dx}{x+1}-\frac13\int\dfrac{x-2}{x^2-x+1}dx.$$

The first integral is obvious. Then

$$\frac{x-2}{x^2-x+1}=\frac12\frac{2x-1-3}{x^2-x+1}=\frac12\frac{(x^2-x+1)'}{x^2-x+1}-\frac32\frac1{x^2-x+1}.$$

Finally,

$$\frac1{x^2-x+1}=\frac{\dfrac43}{\left(\dfrac{2x-1}{\sqrt3}\right)^2+1}$$ which integrates using an arc tangent.

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