Let $a$ be a non zero real number. Evaluate the integral $\int \frac{-7x}{x^{4}-a^{4}}dx$

Question

I hit a wall on this question. Below are my steps

$$\int \frac{-7x}{x^{4}-a^{4}}dx=-7\int \frac{x}{x^{4}-a^{4}}dx$$

Let $u=\frac{x^2}{2}, dx = \frac{du}{x}, x^{4}=4u^{2}.$

$$-7\int \frac{1}{4u^{2}-a^{4}}du=-7\int \frac{1}{(2u+a^2)(2u-a^2)}du$$

Use partial fraction decomposition, $$\frac{1}{(2u+a^2)(2u-a^2)}=\frac{A}{2u+a^{2}}+\frac{B}{2u-a^{2}}.$$

Solve for $A$ and $B$:

$$\begin{cases} A=\frac{1}{-2a^{2}} \\ B=\frac{1}{2a^{2}} \end{cases}$$

Now $$\int \frac{1}{(2u+a^2)(2u-a^2)}du=\int \frac{1}{-2a^2(2u+a^{2})}+\int \frac{1}{2a^{2}(2u-a^{2})}$$ Factoring out $a$ yields $$\frac{7}{2a^{2}}(\int \frac{1}{2u+a^{2}}-\int \frac{1}{2u-a^{2}})$$

Evaluate the integral and substitute $u=\frac{x^{2}}{2}$ back.

My final answer is $$\frac{7}{2a^{2}}(\log(x^2+a^2)-\log(x^2-a^2)).$$

Feedback says my answer is wrong. Where did I mess up?

Answer 1

Mistake appeared in evaluating: $$\int\frac{1}{2u\pm a^2}\,\mathrm{d}u.$$ To solve this you have to substitute $v=2u\pm a^2$, $\mathrm{d}u=1/2 \mathrm{d}v$. We get $$\frac{1}{2}\int\frac{1}{v}\,\mathrm{d}v.$$ From now on everything should be fine.

---

There is also a much easier way to evaluate this integral.

Substitute $u=-ix^2/{a^2}$. The differential is $\mathrm{d}x=ia^2/x^2\,\mathrm{d}u$. $$\int-\frac{7ia^2}{2(-a^4u^2-a^4)}\,\mathrm{d}u = \frac{7i}{2a^2}\int\frac{1}{u^2+1}\,\mathrm{d}u = \frac{7i}{2a^2}\arctan u+C.$$ Substituting back gives us $$\frac{7i\arctan\left(-\frac{ix^2}{a^2}\right)}{2a^2}+C=\frac{7i\operatorname{arctanh}\left(\frac{x^2}{a^2}\right)}{2a^2}+C.$$

If you want to get a solution with logarithm function involved, you can do that with this identity $$\arctan(x)= \frac{1}{2i}\log \left( \frac{x-i}{x+i}\right).$$

Answer 2

You are right up to

$$\frac{7}{2a^{2}}(\int \frac{du}{2u+a^{2}}-\int \frac{du}{2u-a^{2}}) = \frac7{4a^2}\ln\frac{2u+a^2}{|2u-a^2|}= \frac7{4a^2}\ln\frac{x^2+a^2}{|x^2-a^2|} $$

Answer 3

$$I=\int\frac{-7x}{x^4-a^4}dx=-7\int\frac{x}{x^4-a^4}dx$$ now for ease lets let $u=x^2\Rightarrow dx=\frac{du}{2x}$ and $b=a^2$ so: $$I=-\frac72\int\frac{du}{u^2-b^2}$$ now let $u=bv\Rightarrow du=b\,dv$ so: $$I=-\frac{7b}{2}\int\frac{dv}{b^2(v^2-1)}=-\frac{7}{2b}\int\frac{dv}{v^2-1}$$ now this is a standard integral that you can solve with PFD or using a substitution