I've got this question. $$\int \frac{1}{\sqrt{x^2+x}}dx $$ I must use the following substitution $$u=x+\sqrt{x^2+2}$$ What I've done $dx=\frac{1}{\frac{x}{\sqrt{x^2+2}}}du$ which i can put in the integral. How do i continue?
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Do you have to use that substitution? You could use $x=\sinh^2 t$ instead – Ninad Munshi Oct 25 '19 at 21:23
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@user9060784 Note that the person who asked can mark one answer as "accepted". Please visit https://math.stackexchange.com/tour – Robert Z Oct 26 '19 at 09:08
2 Answers
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The integration is easily carried out if you use the substitution $$u=x+\frac{1}{2}.$$
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There is a mistake in the proposed substitution, it should be $u=x+\sqrt{x^2+x}$
By squaring $(u-x)$ and solving for $x$ you have $x=\dfrac{u^2}{2u+1}$
And $x^2+x=\dfrac{u^2(u+1)^2}{(2u+1)^2}$
Note: I ommitted absolute values in the calculations, considering $x\ge 0\implies u\ge 0$, but in the other interval $x\le -1$ makes $u\le 0$, and this might need some attention. The resulting formula may not be the same exactly.
Then the integral becomes
$$\int\dfrac{dx}{\sqrt{x^2+x}}=2\int \dfrac{\sqrt{\dfrac{u^2(u+1)^2}{(2u+1)^2}}}{u(u+1)}\mathop{du}=2\int\dfrac{du}{2u+1}=\ln(2u+1)+C$$
And by subtracting $\ln(2)$ from the constant this gives: $$\ln\left(\frac 12+x+\sqrt{x^2+x}\right)+D=\ln\left((x+\frac 12)+\sqrt{(x+\frac 12)^2-\frac 14}\right)+D$$
Exposing why the symmetric change proposed by S.Dolan in his answer is also pertinent.
zwim
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