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Hello I need some help with the following integral:

$$\int\frac{dx}{x\sqrt{x^2+x+1}}$$

Have been trying u-sub, and parts which do not get me to a solution!

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2 Answers 2

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If we use the Euler substitution $\sqrt{x^2+x+1}=t-x$, so $\displaystyle x=\frac{t^2-1}{2t+1}dt$ and $\displaystyle dx=\frac{2(t^2+t+1)}{(2t+1)^2}dt$,

we get $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{\frac{t^2-1}{2t+1}\left(t-\frac{t^2-1}{2t+1}\right)}\frac{2(t^2+t+1)}{(2t+1)^2}dt=2\int\frac{1}{t^2-1}dt$.

Using partial fractions, this gives

$\displaystyle2\int\left(\frac{1/2}{t-1}-\frac{-1/2}{t+1}\right)dt=\ln|t-1|-\ln|t+1|+C$

$=\ln\left|\sqrt{x^2+x+1}+x-1\right|-\ln\left|\sqrt{x^2+x+1}+x+1\right|+C$.


Alternate Solution:

If we use the Euler substitution $\sqrt{x^2+x+1}=tx+1$, so $\displaystyle x=\frac{2t-1}{1-t^2}$ and $\displaystyle dx=\frac{2(t^2-t+1)}{(1-t^2)^2} dt$,

we get $\displaystyle\int\frac{1}{x\sqrt{x^2+x+1}}dx=\int\frac{1}{\frac{2t-1}{1-t^2}\cdot\frac{t^2-t+1}{1-t^2}}\frac{2(t^2-t+1)}{(1-t^2)^2} dt=\int\frac{2}{2t-1}dt$

$\;\;\ln|2t-1|+C=\ln\left|2\left(\frac{\sqrt{x^2+x+1}-1}{x}\right)-1\right|+C=\ln\left(2\sqrt{x^2+x+1}-2-x\right)-\ln|x|+C$.

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Complete the square to get $\dfrac{1}{x\sqrt{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}}$ and substitute $u=x+\dfrac{1}{2}$

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  • $\begingroup$ This was my first thought, but it doesn't lead to that nice of a problem. The answer also contains $\ln(x)$, so it is not that clear why this is the correct approach. $\endgroup$
    – Ian
    Commented Sep 22, 2014 at 22:32

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