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Integrate $\int\frac{2t^2}{t^4+1}dt$

While evaluating the integral $\int\sqrt{\tan x}dx$ in Evaluating the indefinite integral $\int\sqrt{\tan x}dx$. using the substitution $t^2=\tan x\implies2tdt=\sec^2x.dx$, thus $$ \int\sqrt{\tan x}dx=\int\frac{2t^2}{t^4+1}dt $$ This is solved using partial fractions, Check answers of @Bhaskara-III, @Harish Chandra Rajpoot. But, what if I try the following

My Attempt $$ \int\frac{2t^2}{t^4+1}dt=\int\frac{2t^2}{(t^2+i)(t^2-i)}dt\\ \frac{2t^2}{(t^2+i)(t^2-i)}=\frac{A}{t^2+i}+\frac{B}{t^2-i}\\ 2t^2=A(t^2-i)+B(t^2+i)\implies A=1, B=1\\ \color{red}{\frac{2t^2}{(t^2+i)(t^2-i)}=\frac{1}{t^2+i}+\frac{1}{t^2-i}}\\ $$ $$ \int\frac{2t^2}{t^4+1}dt=\int\frac{1}{t^2+i}dt+\int\frac{1}{t^2-i}dt=\int\frac{1}{t^2+(\sqrt{i})^2}dt+\int\frac{1}{t^2-(\sqrt{i})^2}dt $$ Is it possible to somehow finish the integration with my substitution ?

Pls check: integrating square root of $\tan x$, answer by @Mhenni Benghorbal, seems to be a similar substitution as in my attempt.

SOORAJ SOMAN
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  • It doesn't really make sense to introduce $i$ when you're in the real domain... – TheSimpliFire May 27 '18 at 14:12
  • This is a duplicate question, it has been asked before and contains a very extensive solution –  May 27 '18 at 14:13
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    @James this question is not entirely directed at integrating $\sqrt{\tan x}$, my doubt is regarding simplifying partial fraction with introducing $i$. – SOORAJ SOMAN May 27 '18 at 14:16
  • @James Alright. I have modified OP a bit, hope this 'd make my doubt clear. – SOORAJ SOMAN May 27 '18 at 14:24
  • If you want continue using this way, continue the partial fraction decomposition. – Claude Leibovici May 27 '18 at 15:29
  • @ClaudeLeibovici could u pls elaborate on that ?. – SOORAJ SOMAN May 27 '18 at 15:45
  • @ss1729 I think you can do it this way. But the complex numbers can ruin it, mostly because the $\exp$ function's (formal) inverse is not a function. – Botond May 27 '18 at 16:42
  • If you want to continue with complex numbers you can do the following. Let $\xi=(1+i)/\sqrt2$. Then the zeros of $t^4+1$ are $\xi^j, j=1,3,5,7$. You can find a partial fraction decomposition of the form $$\frac{2t^2}{t^4+1}=\frac A{t-\xi}+\frac B{t-\xi^3}+\frac C{t-\xi^5}+\frac D{t-\xi^7}.$$ The indefinite integral you get involves complex logarithms, and you need to be familiar with those. Here $\xi$ and $-\xi=\xi^5$ are the square roots of $i$, and $\pm\xi^3$ are the square roots of $-i$. – Jyrki Lahtonen May 27 '18 at 17:12
  • (cont'd) So the $A$ and $C$ terms should add up to your $1/(t^2-i)$ and the $B$ and $D$ terms give your $1/(t^2+i)$. – Jyrki Lahtonen May 27 '18 at 17:14
  • @JyrkiLahtonen thnx. but how do I further take it from there ?.could u pls help ? – SOORAJ SOMAN May 27 '18 at 18:29
  • If $z$ is one of those $\xi^j$, and $A_j$ is the corresponding numerator (so $A=A_1, B=A_3, C=A_5, D=A_7)$ then, because the zeros are simple, $$A_j=\lim_{t\to z}\frac{2t^2(t-z)}{t^4+1}=\lim_{t\to z}\frac{4t(t-z)+2t^2}{4t^3}=\frac1{2z}$$ by l'Hospital. So $A=-C=(1-i)/(2\sqrt2)$ and $D=-B=(1+i)/(2\sqrt2)$. Then it becomes a linear combination of complex logarithms. – Jyrki Lahtonen May 27 '18 at 18:54
  • Jyrki Lahtonen answered your question. – Claude Leibovici May 28 '18 at 03:16
  • Yes, it is indeed possible to finish off the given integral with your substitution, albeit the process is tedious as we have to invoke the logarithmic definitions of inverse trigonometric functions and the definition of the complex logarithm so that the final anti-derivative is a real-valued function. I don't see why everyone is objecting to this. – Integreek Feb 07 '25 at 11:59

5 Answers5

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I may give another way.

From $$ \int \frac{2t^{2}}{1+t^{4}} dt $$ Divide by $t^2$ $$ \int \frac{2}{\frac{1}{t^{2}}+t^{2}} dt $$ $$ \int \frac{1 + 1/t^{2} + 1 - 1/t^{2}}{\frac{1}{t^{2}}+t^{2}} dt $$ $$ \int \frac{1 + t^{-2}}{\frac{1}{t^{2}}+t^{2}} dt + \frac{1 - t^{-2}}{\frac{1}{t^{2}}+t^{2}} dt $$ Then let $a=(t - \frac{1}{t})$ for left and $b= (t + \frac{1}{t})$ for right part to continue.

P.S. I saw this in students' assignment solution

Redsbefall
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Hint

You have $$\int \frac {2t^2dt}{t^4+1}$$

Which can be written as $$\int \frac {(t^2+1+t^2-1)dt}{t^4+1}$$

$$\int \frac {(t^2+1)dt}{t^4+1}+\int \frac {(t^2-1)dt}{t^4+1}$$

$$\int \frac {(1+t^{-2})dt}{t^{-2}+t^2}+\int \frac {(1+t^{-2})dt}{t^2+t^{-2}}$$

For the first one substitute $$u=x-\frac 1x$$ and for second one substitute $$p=x+\frac 1x$$

Rohan Shinde
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$$t^4+1=(t^2+\sqrt2t+1)(t^2-\sqrt2t+1)$$

By expanding in partial fraction:

$$\frac{2t^2}{1+t^4}=\frac{t/\sqrt2}{t^2-\sqrt2t+1}-\frac{t/\sqrt2}{t^2+\sqrt2t+1}$$

To find the partial fraction decomposition, write

$$\frac{2t^2}{1+t^4}=\frac{at+b}{t^2+\sqrt2t+1}+\frac{ct+d}{t^2-\sqrt2t+1}$$

Multiply both sides by $1+t^4$ and simplify:

$$2t^2=(a+c)t^3+(b+d-\sqrt2a+\sqrt2c)t^2+(a+c+\sqrt2d-\sqrt2b)t+(b+d)$$

Now identify coefficients:

  • from $t^3$, $a+c=0$
  • from the constant, $b+d=0$
  • from $t$ (using $a+c=0$): $b=d$, then $b=d=0$ since $b+d=0$
  • from $t^2$ (using $b+d=0$), $2=\sqrt2(c-a)=2c\sqrt2$, hence $c=\frac{\sqrt2}2$
  • since $a=-c$, $a=-\frac{\sqrt2}{2}$

The integral of the first term leads to:

$$\int\dfrac{t/\sqrt2}{t^2-\sqrt2t+1}\mathrm dt=\frac{\sqrt2}{4}\int\dfrac{2t-\sqrt2+\sqrt2}{t^2-\sqrt2t+1}\mathrm dt\\=\frac{\sqrt2}{4}\log|t^2-\sqrt2t+1|+\frac12\int\dfrac{1}{t^2-\sqrt2t+1}\mathrm dt$$

And

$$\int\dfrac{\mathrm dt}{t^2-\sqrt2t+1}=\int\dfrac{\mathrm dt}{\left(t-\frac{\sqrt2}{2}\right)^2+\frac12}=2\int\dfrac{\mathrm dt}{\left(\sqrt2t-1\right)^2+1}$$

With the change of variable $u=\sqrt2t-1$, this integral is

$$2\int\dfrac{\mathrm dt}{\left(\sqrt2t-1\right)^2+1}=\sqrt2\int\frac{\mathrm du}{u^2+1}=\sqrt2\arctan(u)+C=\sqrt2\arctan(\sqrt2t-1)+C$$

The first term thus yields:

$$\int\dfrac{t/\sqrt2}{t^2-\sqrt2t+1}\mathrm dt=\frac{\sqrt2}{4}\log|t^2-\sqrt2t+1|+\frac{\sqrt2}{2}\arctan(\sqrt2t-1)+C$$

The second term is similar:

$$\int\dfrac{t/\sqrt2}{t^2+\sqrt2t+1}\mathrm dt=\frac{\sqrt2}{4}\int\dfrac{2t+\sqrt2-\sqrt2}{t^2+\sqrt2t+1}\mathrm dt\\=\frac{\sqrt2}{4}\log|t^2+\sqrt2t+1|-\frac12\int\dfrac{1}{t^2+\sqrt2t+1}\mathrm dt$$

And

$$\int\dfrac{\mathrm dt}{t^2+\sqrt2t+1}=\int\dfrac{\mathrm dt}{\left(t+\frac{\sqrt2}{2}\right)^2+\frac12}=2\int\dfrac{\mathrm dt}{\left(\sqrt2t+1\right)^2+1}$$

With the change of variable $u=\sqrt2t+1$, this integral is

$$2\int\dfrac{\mathrm dt}{\left(\sqrt2t+1\right)^2+1}=\sqrt2\int\frac{\mathrm du}{u^2+1}=\sqrt2\arctan(u)+C=\sqrt2\arctan(\sqrt2t+1)+C$$

The second term thus yields:

$$\int\dfrac{t/\sqrt2}{t^2+\sqrt2t+1}\mathrm dt=\frac{\sqrt2}{4}\log|t^2+\sqrt2t+1|-\frac{\sqrt2}{2}\arctan(\sqrt2t+1)+C$$

All in all

$$\int\dfrac{2t^2\mathrm dt}{1+t^4}=\frac{\sqrt2}{4}\log\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+\frac{\sqrt2}2\arctan(\sqrt2t-1)+\frac{\sqrt2}2\arctan(\sqrt2t+1)+C$$

You can remove the absolute value, as the numerator and the denumerator are both positive:

$$\int\dfrac{2t^2\mathrm dt}{1+t^4}=\frac{\sqrt2}{4}\log\left(\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right)+\frac{\sqrt2}2\arctan(\sqrt2t-1)+\frac{\sqrt2}2\arctan(\sqrt2t+1)+C$$

Jean-Claude Arbaut
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  • could u pls explain a bit on how u did partial fraction decomposition on $\frac{2t^2}{t^4+1}=\frac{2t^2}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}$ – SOORAJ SOMAN May 27 '18 at 20:18
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    @ss1729 The simplest is to write $\dfrac{2t^2}{1+t^4}=\dfrac{at+b}{t^2+\sqrt2t+1}+\dfrac{ct+d}{t^2-\sqrt2t+1}$, simplify the sum and identify the coefficients $a,b,c,d$. From this you know it has necessarilty this form. – Jean-Claude Arbaut May 27 '18 at 20:26
  • i was trying tht. think its a rather cumbersome process to find the coefficients. – SOORAJ SOMAN May 27 '18 at 20:28
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    @ss1729 You can simplify a bit. Let $t=0$ in both and you find that $b+d=0$. Multiply by $t^2$ then let $t\to\infty$ and you find that $a+c=0$. With two more value you would have a $2\times2$ linear system to solve. – Jean-Claude Arbaut May 27 '18 at 20:39
  • @ss1729 I updated with the identification of coefficients, so that you can check what you are trying. – Jean-Claude Arbaut May 27 '18 at 20:49
  • @ss1729 Yes, but not exactly that. You are computing the tangent of the sum and you simplify, and that gives $\dfrac{\sqrt{2}t}{1-t^2}$. However, when $t$ is large enough, the sum of the two arctangents is larger than $\pi/2$, and if you take the arctangent of the tangent, you get another value (it's necessarily in $]-\pi/2,\pi/2[$). Thus you can't do that. Also, it may happen that another method of integration leads to another, equivalent formula. It's not always easy to see they are equivalent, and they may differ by a constant. – Jean-Claude Arbaut May 27 '18 at 22:18
  • thank u. got it. it comes from $\tan x+\tan{\frac{1}{x}}=\frac{-\pi}{2}$ for $x<1$. – SOORAJ SOMAN May 27 '18 at 22:20
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To answer your question, it doesn't make much sense to introduce $i$ by factoring the denominator as $t^2+i$ and $t^2-i$, especially since you're finding the integral of $\sqrt{\tan x}$ over the real domain. Of course, it also wouldn't help to try it out.

With that being said, another way to reduce the integral is to divide both the numerator and denominator by $t^2$ and split the integrand into two separate "partial fractions" and integrate each one. You should get an $\arctan(\cdot)$ and a $\log(\cdot)$.

$$\begin{align*}I & =\int dt\,\frac {2}{t^2+t^{-2}}\end{align*}$$

Frank W
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Your approach is quite tedious but doable. Assuming that $\sqrt i$ is $e^\frac{\pi i}4=\frac{i+1}{\sqrt2}$(it can be assumed to be $e^\frac{5\pi i}4$ also),

$$\begin{align}\int\frac{2t^2}{t^4+1}\mathrm dt&=\int\frac{\mathrm dt}{t^2+\left(\sqrt i\right)^2}+\int\frac{\mathrm dt}{t^2-\left(\sqrt i\right)^2}\\&=\frac1{\sqrt i}\tan^{-1}\frac{t}{\sqrt i}+\frac1{2\sqrt i}\log\left(\frac{t-\sqrt i}{t+\sqrt i}\right)+C\\&\overset{(1)}{=}-\frac{\sqrt i}2\log\left(\frac{1+\sqrt it}{1-\sqrt it}\right)+\frac1{2\sqrt i}\log\left(\frac{t-\sqrt i}{t+\sqrt i}\right)+C\\&=\frac{-1-i}{2\sqrt2}\log\left(\frac{\sqrt2+t+it}{\sqrt2-t-it}\right)+\frac{1-i}{2\sqrt2}\log\left(\frac{t\sqrt2-1-i}{t\sqrt2+1+i}\right)+C\\&\overset{(2)}{=}\frac{-1-i}{2\sqrt2}\left[\log(\sqrt2+t+it)-\log(\sqrt2-t-it)\right]+\frac{1-i}{2\sqrt2}\left[\log(t\sqrt2-1-i)-\log(t\sqrt2+1+i)\right]+C\\&=\frac{-1-i}{2\sqrt2}\left[\frac12\ln\left|2t^2+2\sqrt2t+2\right|+i\tan^{-1}\frac{t}{t+\sqrt2}-\frac12\ln\left|2t^2-2\sqrt2t+2\right|-i\tan^{-1}\frac{t}{t-\sqrt2}\right]+\frac{1-i}{2\sqrt2}\left[\frac12\ln\left|2t^2-2\sqrt2+2\right|+i\tan^{-1}\frac1{1-\sqrt2t}-\frac12\ln\left|2t^2+2\sqrt2+2\right|-i\tan^{-1}\frac1{1+\sqrt2t}\right]+C\\&=\frac{-1-i}{2\sqrt2}\left[-\frac12\ln\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+i\tan^{-1}\frac{\frac{t}{t+\sqrt2}-\frac{t}{t-\sqrt2}}{1+\frac{t^2}{t^2-2}}\right]+\frac{1-i}{2\sqrt2}\left[\frac12\ln\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+i\tan^{-1}\frac{\frac1{1-\sqrt2t}-\frac1{1+\sqrt2t}}{1+\frac{1}{1-2t^2}}\right]+C\\&=\frac{-1-i}{2\sqrt2}\left[-\frac12\ln\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+i\tan^{-1}\frac{\sqrt2t}{1-t^2}\right]+\frac{1-i}{2\sqrt2}\left[\frac12\ln\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+i\tan^{-1}\frac{\sqrt2t}{1-t^2}\right]+C\\&=\boxed{\frac1{2\sqrt2}\ln\left|\frac{t^2-\sqrt2t+1}{t^2+\sqrt2t+1}\right|+\frac1{\sqrt2}\tan^{-1}\frac{\sqrt2t}{1-t^2}+C}\end{align}$$

Explanation of steps:

$(1):$ $\tan^{-1}x=-i\tanh^{-1}ix=\dfrac{-i}2\log\dfrac{1+ix}{1-ix}$

$(2):$ $\log z=\ln|z|+i(2n\pi+\text{arg}z), n\in\mathbb Z$

Integreek
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