The integral of $\frac{x}{16-x^4}$

Question

$$\int\frac{x}{16-x^4}\;dx$$ $\because$ $\frac{d}{dx}(\coth^{-1}\frac{x}{a}$)$=\frac{1}{a^2-x^2}$ ,And also $\frac{d}{dx}(\tanh^{-1}\frac{x}{a}$)$=\frac{1}{a^2-x^2}$

$\therefore$ If we say that $u=x^2$ we can rewrite the integral as $$\frac{1}{2}\int\frac{1}{4^2-u^2}\;du$$

Which of the two answers will be correct

$\frac{1}{8}\tanh^{-1}(\frac{x^2}{4})+C$ or $\frac{1}{8}\coth^{-1}(\frac{x^2}{4})+C$

Or are both of them correct and, if so, why ?

Answer 1

$ \newcommand{\arccoth}{\operatorname{arccoth}} \newcommand{\arctanh}{\operatorname{arctanh}} $They're both correct, in a sense.

As you have noticed, $\arctanh(x)$ and $\arccoth(x)$ have the same derivative. However, they're very different functions.

The key issue at play here is that they have totally different domains.

• $\arccoth(x)$ is only defined for $|x|>1$
• $\arctanh(x)$ is only defined for $|x|<1$

They have derivatives described by the same function, but only when restricting the domain to that set. Hence:

• $\arccoth(x)$'s derivative looks like $1/(1-x^2)$ when graphed on $(-\infty,-1)\cup(1,\infty)$
• $\arctanh(x)$'s derivative looks like $1/(1-x^2)$ when graphed on $(-1,1)$

Consequently, for an integral of this form, an antiderivative doesn't tell you the whole story. You need to look at some definite integral over $[a,b]$, and use whichever antiderivative applies for that interval.

Answer 2

The domain of $\operatorname{arctanh}$ is $(-1,1)$, whereas the domain of $\operatorname{arccoth}$ is $(-\infty,-1)\cup(1,\infty)$. So, both answers can be correct. It depends upon the interval with wich you are working.