How $\mathrm dx$ converted to $\mathrm d$?

Question

Find the value of $\int \dfrac{x^2+1}{x^4+1}\mathrm dx$.

$$\int \frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}\mathrm dx$$ $$\int \frac{d(x+\frac{1}{x})}{(x+\frac{1}{x})^2-2}$$

How to integrate 2nd $$\int \frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}\mathrm dx$$ equation? Even, how they had found that $$\int \frac{x^2(1-\frac{1}{x^2})}{x^2(x^2+\frac{1}{x^2})}\mathrm dx =\int \frac{d(x+\frac{1}{x})}{(x+\frac{1}{x})^2-2}$$ Where $\mathrm dx$ had gone in second line? How $\mathrm d$ came left?

I am not doing integration by parts or, technics of integration. I am just doing some standard integrals. I found this kind of 3 problems. How can I solve that simple way?

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Replying to my title : I have read somewhere that they had differentiate inside function? Why they had? What's the reason of differentiating that?

$$\int dx= x +c$$ So, if we differentiate the inside function then, it cancels with integration. So, is it wise to differentiate inside function? While I am differentiate inside function than, I have add another integral, but, that didn't happen here. So, how it is correct?

Answer 1

First of all this is a very standard question in indefinite integrals and this method is the fastest way to solve it. To address your concern,$$\frac{d}{dx}(x+\frac{1}{x})=1-\frac{1}{x^2}$$ When you substitute this in the integral, $dx$ cancels out and hence they directly wrote $d(x+\frac{1}{x})$.
Also to avoid all this, just assume $$x+\frac{1}{x}=t \Rightarrow (1-\frac{1}{x^2})dx=dt$$ An this can be further solved. Remember that whenever you face functions such as, $$\int \frac{x^2\pm 1}{x^4\pm kx^2+1}dx$$ always use the approach used in Original Post. As this method converts the question into an integrable function when you take out $x^2$ as common, you can write $x^2+\dfrac{1}{x^2}=(x+\dfrac{1}{x})^2-2$ and the derivative of $(x+\dfrac{1}{x})$ will be present in numerator and hence it is an integrable form which is $$\frac{dt}{t^2-2}$$

Answer 2

Hint :Use Partial Fration i.e. $${\int{\frac{x^{2} + 1}{x^{4} + 1} d x}} = {\int{\left(\frac{1}{2 \left(x^{2} + \sqrt{2} x + 1\right)} + \frac{1}{2 \left(x^{2} - \sqrt{2} x + 1\right)}\right)d x}}$$

Answer 3

d$\left( x+\frac 1 x\right)$ stands for $\frac d {dx} \left( x+\frac 1 x\right)$. So, that step basically means that we have done a change of variables $x\to\left( x+\frac 1 x\right)$

It is same as substituting $y=\left( x+\frac 1 x\right)$, saying that $dy=x^2\left( 1-\frac 1 {x^2}\right)dx$, writing the whole integrand in terms of $y$, and integrating it.

Does that help?