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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?

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?

  • @gt6989b What's the difference between \frac{}{} and \dfrac{}{}?> –  Jul 19 '21 at 05:05
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    Use Cauchy integral method by integrating in complex plane along a path via +x-axis,+y-axis and a radial segment. – Shreedhar Bhat Jul 19 '21 at 05:06
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    @ShreedharBhat I said I am just doing some standard integrals (simple integral) –  Jul 19 '21 at 05:07
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    @Istiak it's just the difference of size – p_square Jul 19 '21 at 05:11
  • @Koro I guess, I wrote that.. –  Jul 19 '21 at 05:12
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    \frac{a}{b} renders as $\frac{a}{b}$ and dfrac{a}{b} renders as $\dfrac{a}{b}$ in larger scale – gt6989b Jul 19 '21 at 05:36
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    A rule of thumb: Most of the indefinite integrals that occur in exercises have already been covered on this site. ApproachZero is your friend. – Jyrki Lahtonen Jul 19 '21 at 05:41
  • @JyrkiLahtonen No satisfied answer is available there. So, I am pressing on no.. –  Jul 19 '21 at 05:41
  • That comment was autogenerated by the system. In my opinion mostly addressing wannabe-answerers and reviewers. You will see that many users will want to reproduce a standard calculation. – Jyrki Lahtonen Jul 19 '21 at 05:46
  • @JyrkiLahtonen I know, for your information I had wrote that comment. Maybe, new reviewer may flag to that question also. That's the main reason of writing that comment. (Both questions are same. But, I had two question...) –  Jul 19 '21 at 05:48
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    I see. I was chiefly concerned about there coming more answers just calculating the integral by whatever means. – Jyrki Lahtonen Jul 19 '21 at 10:14
  • @gt6989b \displaystyle would have been better here, rather than \dfrac, as \dfrac just alters the fraction, putting it out of proportion with everything else, while \displaystyle makes everything bigger. Compare $\int \dfrac{x^2+1}{x^4+1}\mathrm dx$, via \dfrac, with $\displaystyle\int \frac{x^2+1}{x^4+1}\mathrm dx$, via \displaystyle. – user1729 Jul 19 '21 at 15:28
  • @user1729 agree, in my posts i prefer using $$ instead. – gt6989b Jul 19 '21 at 22:45

3 Answers3

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

p_square
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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 $$\frac{dt}{t^2-2}$$

UNAN
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  • Another question came to my mind, So, I am upvoting instead of accepting as correct answer.... I like your answer.. –  Jul 19 '21 at 06:10
  • Look! I have edited my question... –  Jul 19 '21 at 06:12
  • Sorry, but I am not able to understand your actual concern @Istiak – UNAN Jul 19 '21 at 06:14
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    What did you mean by OP. I don't understand the word. Everyone uses every time they uses for different things. Why did you mean by it? –  Jul 19 '21 at 06:33
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    @Istiak, Original Post – UNAN Jul 19 '21 at 06:35
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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?

Sayan Dutta
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