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I need to solve this integral

$$\int \frac{x-1}{x+4x^3} \, dx$$

As some users suggest me in my previous question here I used the partial fractions.

and I got:

$$\int \frac{x-1}{x+4x^3} \, dx = - \int \frac{1}{x} \, dx + \int \frac{4x+1}{1+4x^2} \, dx$$

The problem is the second integral, I have tried to separate it in this way:

$$\int \frac{4x+1}{1+4x^2} \, dx = \int \frac{4x}{1+4x^2} \, dx + \int \frac{1}{1+4x^2} \, dx$$

and the final result I get is:

$$-\ln|x| + 4\arctan(2x) + \arctan(2x)$$

But the result is wrong, what am I missing?

EDIT: I have fixed the x^2

Leucippus
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Christian Giupponi
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1 Answers1

2

note that $$\frac{x-1}{x+4x^3}=-\frac 1x+\frac{4x+1}{4x^\color{red}2+1}$$ for the second then write

$$\frac{4x+1}{4x^2+1}=\frac12\frac{2}{(2x)^2+1}+\frac12\frac{8x}{4x^2+1}$$

EDIT:

Hence \begin{align} \int \frac{x-1}{x+4x^3}dx&=-\int\frac1x dx+\frac12\int\frac{2}{(2x)^2+1}dx+\frac12\int\frac{8x}{4x^2+1}dx\\ &=-\log x+\frac12\int\frac{d(\tan t)}{\tan^2t+1} \color{blue}{\text{(with } 2x\to\tan t)}+\frac12\log(4x^2+1) \\ &=-\log x+\frac12\int\frac{1+\tan^2 t}{\tan^2t+1}dt +\frac12\log(4x^2+1) \\ &=-\log x+\frac12t +\frac12\log(4x^2+1) \\ &=-\log x+\frac12\arctan(2x) \color{blue}{\text{(with } t\to\arctan 2x)} +\frac12\log(4x^2+1) \\ \end{align}

Math-fun
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