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I want to solve this integral: $$ \int \frac{dx}{x+\sqrt{x^2+x+1}} $$ I converted the quadratic equation into a full squere and got this $$ \int \frac{dx}{x+\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}} $$ then I put x+1/2 = t and got $$ \int \frac{dt}{t-\frac{1}{2}+\sqrt{t^2+\frac{3}{4}}} $$

And I don't know how to continue from here, what would be the next step and are these steps so far good? Thanks.

letsgooo
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    try substitution $t=x+\sqrt{x^2+x+1}$ – Vasili Apr 05 '22 at 15:52
  • the value of dx doesn't make sense to me if i try that – letsgooo Apr 05 '22 at 16:09
  • The indication given by @Vasili is a good one (you have to express $x$ as a function of $t$: $(t-x)^2=x^2+x+1 \implies t^2-2tx-(x+1)=0$ and then extract $x=...$). Another method (more complicated) : multiply numerator and denominator by the so-called conjugated expression of the denominator: $x-\sqrt{x^2+x+1}$. – Jean Marie Apr 05 '22 at 16:12

2 Answers2

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Hint: use $t=x+\sqrt{x^2+x+1} \implies t-x=\sqrt{x^2+x+1} \implies t^2-2tx=x+1$
Thus, $x=\frac{t^2-1}{1+2t}, dx=\frac{2(t^2+t+1)}{(1+2t)^2}$
This will lead to integral of rational function which is usually solved by partial fraction decomposition.

Vasili
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Let $y=\sqrt{t^2+\frac34}$, we have $y^2-t^2=\frac34$, so

$$\int\frac{d(y+t)}{y+t-\frac12}=\log\left|y+t-\frac12\right|+c_+,$$ $$\int\frac{d(y-t)}{y+t-\frac12}=\int\frac{(y-t)d(y-t)}{\frac34-\frac12(y-t)}=-3\log\left|\frac34-\frac12(y-t)\right|-2(y-t)+c_-,$$

Therefore $$\int\frac{dt}{y+t-\frac12}=\frac12\log\left|y+t-\frac12\right|+\frac32\log\left|\frac34-\frac12(y-t)\right|+y-t+c.$$

Yizhen Chen
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