Integrating $\frac{1}{(1-7x^{2})^{2}}$

Question

How do I integrate the following algebraic function?

$$\int \frac{1}{(1-7x^{2})^{2}} \, {\rm d} x$$

Really , without use of hyperbolic function , can it be done ... Actually , my syllabus doesn't allow using hyperbolic functions ...

Answer 1

Hint:

Let

$$I:=\int\frac{dx}{1-x^2}=\text{artanh }x,\\ J:=\int\frac{dx}{(1-x^2)^2}$$

and notice that$^*$

$$\left(\frac x{1-x^2}\right)'=\frac{1+x^2}{(1-x^2)^2}=\frac{2-(1-x^2)}{(1-x^2)^2}$$ so that

$$2J-I=\frac x{1-x^2}.$$

This yields

$$2J=\frac x{1-x^2}-\text{artanh }x.$$

I let you do the easy adjustments to deal with the coefficient $7$.

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$^*$This trick is discovered by observing that the square at the denominator might come from the derivative of $\dfrac{P(x)}{1-x^2}$. If we are lucky, we can find a polynomial $P(x)$ such that the derivative will be of the form $\dfrac{ax^2+bx+c}{(1-x^2)^2}$, because this can be written $A+\dfrac B{1-x^2}+\dfrac{Cx}{1-x^2}$ where all terms are easy to integrate.

Trying $P(x)=1$ and $P(x)=x$ leads us to the solution.

Answer 2

Hint: $$\int\frac{1}{49x^4-14x^2+1}dt={1\over2}\int\frac{2\over x^2}{(7x)^2+{1\over x^2}-14}$$

Now try completing the square and substitution

Answer 3

$$I=\int\frac{1}{(1-7x^2)^2}dx =\int x^{-1}\cdot \frac{x}{(1-7x^2)^2}dx$$

Using Integration by parts method

$$I=\frac{1}{14}\cdot \frac{1}{x(1-7x^2)}-\frac{1}{14}\int\frac{1}{x^2(1-7x^2)}dx$$

$$I=\frac{1}{14}\cdot \frac{1}{x(1-7x^2)}-\frac{1}{14}\int\bigg[\frac{1}{x^2}-\frac{1}{1-7x^2}\bigg]dx$$

$$I=\frac{1}{14}\cdot \frac{1}{x(1-7x^2)}+\frac{1}{x}-\frac{1}{2\sqrt{7}}\ln\bigg|\frac{1+\sqrt{7}x}{1+\sqrt{7}x}\bigg|+\mathcal{C}$$