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First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:

$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$ for this I have that $A=0, B=2, C=0, D=-2$

so now I have $I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$

Now,

$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$

and

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

using partial integration we have:

$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$

and $$dt=dv \Rightarrow t=v$$

so now we have:

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

so, the final solution should be:

$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$

since the original variable was $x$ we have

$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$

But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$

I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?

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  • $\begingroup$ Did you differentiate both "solutions"? You must get the original integrated function. $\endgroup$
    – DonAntonio
    Commented Feb 27, 2016 at 9:35
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    $\begingroup$ I guess there is a mistake in $du$... $\endgroup$
    – Hugo
    Commented Feb 27, 2016 at 9:39
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    $\begingroup$ You did $\;du\;$ wrong:$$\left(\frac1{(1+t^2)^2}\right)'=-\frac{4t}{(1+t^2)^3}$$ $\endgroup$
    – DonAntonio
    Commented Feb 27, 2016 at 9:44
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    $\begingroup$ @cdummie You're welcome. Comebody seems to have downvoted all, or almost all, the answers. I'd like to know why. $\endgroup$
    – DonAntonio
    Commented Feb 27, 2016 at 9:47
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    $\begingroup$ @Joanpemo exactly, the more answers i get, more ways and tricks for solving i have, which can be useful not only for this example, but for some other examples. $\endgroup$
    – cdummie
    Commented Feb 27, 2016 at 9:57

6 Answers 6

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You have mistake here:

$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{\color{red}(1+t^2\color{red}{)^3}}$

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I'd try the following: substitute $\;t^2=x\implies 2t\,dt=dx\;$, and your integral becomes

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

and already here integrate by parts:

$$\begin{cases}u=t&u'=1\\{}\\v'=\frac{2t}{(t^2+1)^2}&v=-\frac1{t^2+1}=\end{cases}\;\;\;\implies$$

$$I=-\frac t{t^2+1}+\int\frac1{1+t^2}dt=-\frac t{1+t^2}+\arctan t+C$$

and going back to the original variable

$$I=-\frac{\sqrt x}{x+1}+\arctan\sqrt x+C$$

so I think the book's right.

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Alternative Approach:

Let $x=\tan^2{\theta}$,$dx=2\tan{\theta} \sec^2{\theta} d\theta$

\begin{align} I&=\int{\frac{\tan{\theta}\cdot 2\tan{\theta}\sec^2{\theta} d\theta}{ \sec^4{\theta} }}\\&=2\int{\sin^2{\theta}}d\theta\\&=\int{1-\cos{(2\theta)}}d\theta\\&=\theta-\frac12 \sin{(2\theta)}+C\\&=\arctan{\sqrt x}-\frac{\sqrt x}{1+x}+C \end{align}

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Integrating by parts, $$\int\dfrac{2t^2}{(1+t^2)^2}dt=t\int\dfrac{2t}{(1+t^2)^2}dt-\int\left(\dfrac{dt}{dt}\cdot\int\dfrac{2t}{(1+t^2)^2}dt\right)dt$$

$$=-\dfrac t{(1+t^2)}-\int\dfrac{dt}{1+t^2}=?$$


Alternatively, choose $\sqrt x=\arctan u\implies x=(\arctan u)^2,dx=\dfrac{\arctan u}{1+u^2}du$

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One may observe that $$ \int \frac{2t}{\left(1+t^2\right)^2} \, dt=\int \frac{\left(1+t^2\right)'}{\left(1+t^2\right)^2} \, dt=-\frac{1}{ 1+t^2} $$ then, using an integration by parts, one has $$ \int \frac{2t^2}{\left(1+t^2\right)^2} \, dt=t \times \left(-\frac{1}{ 1+t^2} \right)+\int \frac1{\left(1+t^2\right)} \, dt=-\frac{t}{ 1+t^2}+\arctan t+C $$

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Notice, you can continue from here without using partial fractions $$\int \frac{2t^2}{(1+t^2)^2}\ dt$$ $$=\int \frac{2t^2}{t^4+2t^2+1}\ dt$$ $$=\int \frac{2}{t^2+\frac{1}{t^2}+2}\ dt$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right)+\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}+2}\ dt$$ $$=\int \frac{\left(1+\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}+\int \frac{\left(1-\frac{1}{t^2}\right)dt}{t^2+\frac{1}{t^2}+2}$$ $$=\int \frac{d\left(t-\frac{1}{t}\right)}{\left(t-\frac{1}{t}\right)^2+4}+\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^2}$$ $$=\frac 12\tan^{-1}\left(\frac{t-\frac 1t}{2}\right)-\frac{1}{\left(t+\frac{1}{t}\right)}+c$$ $$=\frac 12\tan^{-1}\left(\frac{x-1}{2\sqrt x}\right)-\frac{\sqrt x}{x+1}+c$$

let $\sqrt x=\tan\theta\implies \theta=\tan^{-1}(\sqrt x)$, $$=\frac 12\tan^{-1}\left(\frac{\tan^2\theta-1}{2\tan\theta}\right)-\frac{\sqrt x}{x+1}+c$$ $$=\frac 12\tan^{-1}\left(\tan\left(\frac{\pi}{2}+2\theta\right)\right)-\frac{\sqrt x}{x+1}+c$$ $$=\frac{\pi}{4}+\theta-\frac{\sqrt x}{x+1}+c$$

$$=\theta-\frac{\sqrt x}{x+1}+C$$ $$=\color{red}{\tan^{-1}(\sqrt x)-\frac{\sqrt x}{x+1}+C}$$

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    $\begingroup$ could someone please explain the reason for what is wrong in the answer? $\endgroup$ Commented Feb 27, 2016 at 10:05
  • $\begingroup$ I didn't downvote, but your method seems quite long, compared to Joanpemo's $\endgroup$
    – egreg
    Commented Feb 27, 2016 at 10:11
  • $\begingroup$ @egreg: if you please notice you will find the steps are more but method is not so long. I have shown every step. However, making partial fractions is also time taking which is done by OP. $\endgroup$ Commented Feb 27, 2016 at 10:17
  • $\begingroup$ On the other hand, all answers but were downvoted. $\endgroup$
    – egreg
    Commented Feb 27, 2016 at 10:20

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