Evaluate the intgral $\int{\frac{dx}{x^2(1-x^2)}}$ (solution verification)

Question

I have to solve the following integral $$\int{\frac{dx}{x^2(1-x^2)}}$$

What I've got: \begin{split} \int{\frac{dx}{x^2(1-x^2)}} &=\int{\frac{(1-x^2+x^2)dx}{x^2(1-x^2)}}\\ &=\int{\frac{dx}{x^2}}+\int{\frac{dx}{1-x^2}}\\ &=\int{\frac{dx}{x^2}}+\int{\frac{dx}{(1+(xi)^2)}}\\ &=-x^{-1}+\arctan{xi}+C \end{split}

Is this correct?

Thanks in advance!

Answer 1

Don't forget $\dfrac1i$ before $\arctan$ $$\int{\frac{dx}{x^2(1-x^2)}}=-x^{-1}+\dfrac{1}{i}\arctan{xi}+C$$ Also better to write $$\int\dfrac{1}{1-x^2}dx=\dfrac12\ln\dfrac{1+x}{1-x}+C$$

Answer 2

Are you sure that a non-real answer is acceptable? Even if it is, what you should have got as a primitive of $\frac1{1-x^2}$ should have been $-i\arctan(xi)$. But I suggest that you do$$\frac1{1-x^2}=\frac{1/2}{1-x}+\frac{1/2}{1+x}$$instead.

Answer 3

$x = \tanh(u)$ then $dx = \operatorname{sech}^2(u) du$ $$ \begin{align} \int \frac{dx}{x^2(1-x^2)} &= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(1-\tanh^2(u))}\\ &= \int \frac{\operatorname{sech}^2(u) du}{\tanh^2(u)(\operatorname{sech}^2(u))}\\ &= \int \frac{du}{\tanh^2(u)}\\ &= \int \frac{\cosh^2(u)}{\sinh^2(u)} du\\ &= \int (1 + \operatorname{csch}^2(u))du\\ &= u - \coth(u)\\ &= \operatorname{arctanh}(x) - \frac{1}{x} + C \end{align} $$

Answer 4

Note that, if $f(x)=\arctan(xi)$, then $$ f'(x)=i\frac{1}{1+(xi)^2}=\frac{i}{1-x^2} $$ that's not the required derivative; you are missing $1/i=-i$, so you could write $$ -\frac{1}{x}-i\arctan(xi) $$

---

I'd prefer to avoid complex functions. With partial fractions, we set $$ \frac{1}{x^2(1-x)^2}= \frac{a}{x}+\frac{b}{x^2}+\frac{c}{x-1}+\frac{d}{x+1} $$ Removing the denominators leads to $$ ax(x^2-1)+b(x^2-1)+cx^2(x+1)+dx^2(x-1)=-1 $$ With $x=0$, we get $-b=-1$; with $x=1$, we get $2c=-1$; with $x=-1$ we get $-2d=-1$. It remains to find $a$, but this is easily seen to be $0$.

Therefore $$ \int\frac{1}{x^2(1-x^2)}\,dx= \int\left( \frac{1}{x^2}-\frac{1}{2}\frac{1}{x-1}+\frac{1}{2}\frac{1}{x+1} \right)\,dx=-\frac{1}{x}+\frac{1}{2}\log\left|\frac{x+1}{x-1}\right|+c $$

Answer 5

Another method using the self-similar substitution

$$x=\frac{1-t}{1+t}$$

transforms the integral to $$-\int{\frac{2}{{{\left( t-1 \right)}^{2}}}+\frac{1}{2t}dt}$$