How do I solve the integral: $\int \frac{dx}{[x(x^4-4)^\frac{1}{2} ]}$ using substitution method?

Question

I was asked to solve the integral $$\int \frac{dx}{x(x^4-4)^\frac{1}{2}}$$ by using substitution. I've tried many values but none comes to work. I also tried factoring $x^4$ into $x^2\cdot x^2$ in order to try using trigonometric substitution but I could not go anywhere with $\sqrt{x^2\cdot x^2-4}$ . I have no clue what should I do.

I am new to this forum so I am sorry if I broke a rule.

Answer 1

This problem is quite tricky, involving multiple substitutions.

Let $u=x^4-4$. Then $du=4x^3\,dx=\implies dx=\frac{du}{4x^3}$. Substituting that into the integral gives: \begin{align*}\int\frac{dx}{x\sqrt{x^4-4}} &=\int\frac{du}{4x^3(x)\sqrt u} \\ &=\int\frac{du}{4(u+4)\sqrt u} \\ &=\frac14\int\frac{du}{(u+4)\sqrt u}.\end{align*} Now, let $v=\frac{\sqrt{u}}2$. This gives $\displaystyle\int\frac1{v^2+1}$, which is $\arctan(v)$. We substitute back all the variables to get $\boxed{\frac{\arctan\left(\frac{\sqrt{x^2-4}}2\right)}4+C}$. I skipped some intermediate steps, but you should be able to figure them out.

Once you have more experience with these types of problems, you'll gain "intuition" on what to substitute.

Answer 2

HINTS: You start from $$I=\int \frac{dx}{x(x^4-4)^\frac{1}{2}}$$ You mentioned that you want a trigonometric substitution (things like sine and cosine). So you would need to transform your square root into something like $\sqrt{1-t^2}$. How to do this? Take $x^4$ as a factor in the square root. Then $$I=\int\frac{dx}{x\sqrt{x^4\left(1-\frac4{x^4}\right)}}=\int\frac{dx}{x^3\sqrt{1-\left(\frac{2}{x^2}\right)^2}}$$ Can you take it from here?

Answer 3

Hint:

$I = \int \frac{dx}{x \sqrt{x^4 - 4}} = \frac 12 \int \frac{dx}{x \sqrt{\frac {x^4}4 - 1}}$

If $\frac{x^4}4$ were equal to $\sec^2 θ$, we could have simplified the square root as $\tan θ$.

So let's substitute $x = \sqrt{2\sec θ}$ and proceed...

Then we get $dx = \frac1{\sqrt 2}\sqrt{\sec θ} \tanθ \ dθ $

$I = \frac12 \int \frac{\sqrt{\sec θ}\tanθ \ dθ}{\sqrt2\sqrt{2 \secθ} \tanθ} \\ = \frac 14 \int dθ \\ = \frac θ4 \\ = \frac14 \sec^{-1}(\frac {x^2}2) \\ = \frac14 \cos^{-1}(\frac 2{x^2}) + \color{grey}{c}$

Answer 4

$$ \begin{aligned} \int \frac{d x}{x \sqrt{x^4-4}}&=\frac{1}{2} \int \frac{1}{x^4} d\left(\sqrt{x^4-4}\right) \\ &=\frac{1}{2} \int \frac{d\left(\sqrt{x^4-4}\right)}{\left(\sqrt{x^4-4}\right)^2+4} \\ &=\frac{1}{4} \tan ^{-1}\left(\frac{\sqrt{x^4-4}}{2}\right)+C \end{aligned} $$

Answer 5

Substitute $t=\frac 2{x^2}$. Then $$\frac{dx}x=-\frac{dt}{2t},\>\>\>\>\> \frac1{\sqrt{x^4-1}}=\frac t{2\sqrt{1-t^2}}$$ and $$\int \frac{d x}{x \sqrt{x^4-4}}=-\frac14\int \frac{dt}{\sqrt{1-t^2}}=-\frac14\sin^{-1}t+C $$