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While I know that the answer is $-\frac{1}{8}\arctan \frac{4}{\sqrt{x^4-16}},$ I couldn't reach it. When I solved it I reached a different answer, if someone can point at what point I made a mistake and what is the way to solve it I will be thankful.

I will start with taking $u=x^2$, so $du =2x \;dx,$ and $dx = du/2x.$ So I will multiple the integral by $2$ making it $2\int \frac{du}{\sqrt{u^2-4^2}}$ after that I will use the $\cosh^{-1}x$ to make it $2\cosh^{-1}(x^2/4) + c$

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    Do you mean$$\int\frac{dx}{\sqrt[x]{x^4-16}}$$? – 2017-01-16
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    is that $$\int \frac{1}{x\sqrt{x^4-16}}dx$$? – 2017-01-16
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    Yeah it is x * root not power – 2017-01-16
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    What is your question? – 2017-01-16
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    the question is how to solve this integral – 2017-01-16
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    My question : is my answer correct if no in what step did i make mistake and how do the integral can be solved – 2017-01-16
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    If you want to know if your answer is correct, find its derivative either directly or in Wolfram Alpha. – 2017-01-16
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    I know that my answer is wrong but I cant tell at what step I made a mistake – 2017-01-16

2 Answers 2

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$\int \frac{1}{x\sqrt{x^4 - 16}} dx$

Multiply and dividide by $x^3$

$\int \frac{x^3}{x^4 \sqrt{x^4 - 16}} dx$

Put $x^4 - 16 = u^2$

$4x^3 dx = 2u du$

$x^3 dx = \frac{1}{2} u du$

On putting in integral,

= $\frac12 \int \frac{u}{(u^2 + 16) \sqrt{u^2}} du$

= $\frac12 \int \frac{u}{u(u^2 + 16)} du$

= $\frac12 \int \frac{1}{(u^2 + 16)} du$

= $\frac12 \int \frac{1}{(u)^2 + (4)^2} du$

= $\frac12 \tan^{-1} \frac{u}{4} + c$

Then replace value of u.

As you want to know how to do according to your method.

We have $x^2 = u$

$2x dx = du$

$dx = \frac{1}{2x} du$

Then from integral we have,

$\int \frac{1}{2x} \cdot \frac{1}{x \sqrt{u^2 - 16}} du$

= $ \frac{1}{2} \int \frac{1}{x^2 \sqrt{u^2 - 16}} du$

= $ \frac{1}{2} \int \frac{1}{u \sqrt{u^2 - 16}} du$

Then solve it.

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    In answer you are missing 2x in denominator. As dx = $\frac{du}{2x}$ – 2017-01-16
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    If still have any doubt please ask. – 2017-01-16
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    Yes but thats why I multiplied the x in denominator by 2 in order for it to be 2∫dx/2xroot(x^4-16) – 2017-01-16
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    Wait I tell you where you are wrong. – 2017-01-16
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    Hope you got it now. But first method is easier than this. – 2017-01-16
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If you substitute $u^2=x^4-16$ you get a standard integral straight away - try it!

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    I know how to do it know thanks to all the answer but can you tell me at what point I broke a rule of math since in my old way I have reached a different answer. – 2017-01-16
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    You just got the algebra wrong. Substituting $u=x^2$ with $du=2x\,dx$ gives$$\int\frac{1}{x\sqrt{u^2-16}}\frac{du}{2x}=\frac12\int\frac{du}{u\sqrt{u^2-16}}\quad\hbox{not}\quad 2\int \frac{du}{\sqrt{u^2-4^2}}\ .$$ – 2017-01-16