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This is how i solved it:

first i used substitution $x+1 = t^2 \Rightarrow dx=2tdt$

so integral becomes $I=\int \frac{t+2}{t^4-t}2tdt = 2\int \frac{t+2}{t^3-1}dt =2\int\frac{t+2}{(t-1)(t^2-t+1)}dt $

usic partial fraction decomposition i have:

$\frac{t+2}{(t+1)(t^2-t+1)}=\frac{A}{t+1} + \frac{Bt+C}{t^2-t+1}=\frac{At^2-At+A+Bt^2+Bt+Ct+C}{(t+1)(t^2-t+1)}$

from here, we have that $A=\frac{1}{3} , B=-\frac{1}{3}, C=\frac{5}{3}$

so integral becomes $I=\frac{1}{3} \int \frac{dt}{t+1}-\frac{1}{3}\int\frac{(t-5)dt}{t^2-t+1} = \frac{1}{3}ln|t+1|-\frac{1}{3}I_1 $

Now, for the $I_1$

$I_1=\int\frac{(t-5)dt}{t^2-t+1} = \int\frac{tdt}{t^2-t+1} - \int\frac{5dt}{t^2-t+1}= \frac{1}{2}\int\frac{2t+1-1}{t^2-t+1}dt - 5\int\frac{dt}{t^2-t+1}=\int\frac{2t+1}{t^2-t+1}dt - \frac{9}{2}\int\frac{dt}{t^2-t+1}= ln|t^2-t+1|-\frac{9}{2}I_2$

Now, for $I_2$

$I_2=\int\frac{1}{t^2-t+1}dt= \int\frac{1}{t^2-t+\frac{1}{4} + \frac{3}{4}}dt= \int\frac{1}{(t+\frac{1}{2})^2 + \frac{3}{4}}dt=\frac{4}{3} \int\frac{1}{(\frac{2t+1}{\sqrt{3}})^2 + 1}dt$

Now, we can use substitution:

$\frac{2t+1}{\sqrt{3}}=z \Rightarrow dt=\frac{\sqrt{3} dz}{2}$

So we have:

$I_2=\frac{2\sqrt{3}}{3}\int\frac{dz}{1+z^2} =\frac{2\sqrt{3}}{3}\arctan z $

Now, going back to $I_1$

$I_1=ln|t^2-t+1|-3\sqrt{3} \arctan \frac{2t+1}{\sqrt{3}}$

and if we go back to $I$

$I=\frac{1}{3}ln|t+1|-\frac{1}{3} ln|t^2-t+1|-\sqrt{3} \arctan \frac{2t+1}{\sqrt{3}}$

in terms of $x$

$I=\frac{1}{3}ln|\sqrt{x+1}+1|-\frac{1}{3} ln|x+2-\sqrt{x+1}|-\sqrt{3} \arctan \frac{2\sqrt{x+1}+1}{\sqrt{3}}$

Yet, in my workbook i have a different solution, but i can't find any mistakes here, any help?

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  • 4
    $\begingroup$ Your factorization of $t^3-1$ is incorrect. It should be $(t-1)(t^2+t+1)$ $\endgroup$ – GoodDeeds Feb 28 '16 at 14:04
  • $\begingroup$ @GoodDeeds Oh, so that's the problem, what a stupid mistake, thank you! $\endgroup$ – cdummie Feb 28 '16 at 14:06
  • $\begingroup$ You could also write $t^2+t+1 = (t-a)(t-b)$ where $a$ and $b$ are complex and do partial fractions on that. It makes for an interesting exercise in complex logs. $\endgroup$ – marty cohen Feb 29 '16 at 0:03
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$$\int \frac{\sqrt{x+1}+2}{(x+1)^2 - \sqrt{x+1}}dx$$

Set $s=\sqrt{x+1}$ and $ds=\frac{dx}{2\sqrt{1+x}}$

$$\int\frac{(s+2)2s}{s^4-s}ds=\int\frac{2s^2+4s}{s^4-s}ds=2\int\frac{s+2}{s^3-1}ds\overset{\text{partial fractions}}{=}2\int\frac{-s-1}{s^2+s+1}ds+2\int\frac{ds}{s-1}$$

$$=-\int\frac{2s+1}{s^2+s+1}ds-\int\frac{ds}{s^2+s+1}+2\int\frac{ds}{s-1}$$

set $p=s^2+s+1$ and $dp=(2s+1)ds$

$$=-\int \frac 1 p-\int \frac{ds}{s^2+s+1}ds+2\int\frac{ds}{s-1}$$

$$=-\ln|p|-\int\frac{ds}{\left(s+1/2\right)^2+3/4}+2\int\frac{ds}{s-1}$$

Set $w=s+1/2$ and $dw=ds$

$$=-\ln|p|-\frac 4 3\int\frac{dw}{\frac{4w^2}{3}+1}$$

$$=-\ln|p|-\frac{2\arctan(\frac{2w}{\sqrt 3})}{\sqrt 3}-\ln|p|+2\ln|s-1|+\mathcal C$$

$$=\color{red}{2\ln|1-\sqrt{x+1}|-\ln|x+\sqrt{x+1}+2|-\frac{2\arctan\left(\frac{2\sqrt{x+1}+1}{\sqrt 3}\right)}{\sqrt 3}+\mathcal C}$$

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  • 2
    $\begingroup$ Why not doing directly $s=\sqrt{x+1}$? $\endgroup$ – egreg Feb 28 '16 at 14:58
  • $\begingroup$ @egreg You are definitely right , I edited $\endgroup$ – 3SAT Feb 28 '16 at 20:59

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