How to calculate $\int \sqrt{x^2+x+2} \, dx$? [duplicate]

Question

I have to calculate:

$$\int \sqrt{x^2+x+2} \, dx$$

I have tried Euler substitutions. It lead me to polynomial:

$$\int 2\left(\frac{-t^2 + t -2}{1-2t}\right)^2 \, dx$$

Maybe I made a mistake somewhere but this looks ugly to me. There have to be a better method.

I tried to integrate by parts. It lead me to:

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

Unfortunatelly I can't solve $\int \frac{2x^2+x}{\sqrt{x^2+x+2}} \, dx$.

Could you help me to get few steps further?

Thank you for your time.

Answer 1

Integrate by parts, instead, as follows \begin{align} I&=\int \sqrt{x^2+x+2} \ d(x+\frac12)\\ &=(x+\frac12)\sqrt{x^2+x+2} - \int \frac{(x+\frac12)^2}{\sqrt{x^2+x+2}} \, dx\\ &=(x+\frac12)\sqrt{x^2+x+2} -I+\frac74 \int \frac{1}{\sqrt{x^2+x+2}} \, dx\\ &= \frac12 (x+\frac12)\sqrt{x^2+x+2} +\frac78\sinh^{-1}\frac{2x+1}{\sqrt7}+C \end{align}

Answer 2

Since the discriminant of the (monic) quadratic is $-7 < 0$, and its axis of symmetry is $x = -\frac{1}{2}$, we apply the substitution $$x = \frac{1}{2} \left(\sqrt 7 \sinh t - 1\right), \quad dx = \frac{\sqrt 7}{2} \cosh t,$$ which transforms the integral to $$\frac{7}{4} \int \cosh^2 t \,dt = \frac{7}{4} \int\frac{1}{2}(1 + \cosh 2 t)\,dt = \frac{7}{8} \left(t + \frac{1}{2} \sinh 2 t\right) + C = \frac{7}{8} \left(t + \sinh t \cosh t\right) + C .$$ Rearranging the substitution formula gives $$\sinh t = \frac{2 x + 1}{\sqrt 7},$$ so $$t = \operatorname{arsinh} \frac{2 x + 1}{\sqrt{7}} \qquad \textrm{and} \qquad \cosh t = \sqrt{1 + \sinh^2 t} = \frac{2}{\sqrt{7}}\sqrt{x^2 + x + 2}.$$ Back-substituting then yields $$\frac{7}{8} \operatorname{arsinh} \frac{2 x + 1}{\sqrt 7} + \frac{2 x + 1}{4}\sqrt{x^2 + x + 2} + C$$

Answer 3

$$\int \sqrt{x^2+x+2} \, dx$$ $$=\int \sqrt{(x+\frac{1}{2})^2+\frac{7}{4}} \, dx =I$$ $$ y =\frac{\sqrt{7}}{\sqrt{4}}(x+\frac{1}{2})$$ $$dy = \sqrt{\frac{7}{4}}dx$$

$$ I=\frac{7}{4}\int\sqrt{y^2+1} dy $$ then we will use integration by parts $$ \frac{7}{4}\int\sqrt{y^2+1} dy =\frac{7}{4}y\sqrt{y^2+1}-\frac{7}{4}\int\frac{y^2}{\sqrt{y^2+1}} dy$$ note that $$\frac{y^2}{\sqrt{y^2+1}}=\sqrt{y^2+1}-\frac{1}{\sqrt{y^2+1}} $$ so that $$I=\frac{7}{4}y\sqrt{y^2+1} -I+\frac{7}{4}\int\frac{1}{\sqrt{y^2+1}}dy$$ now we have to figure $$\int\frac{1}{\sqrt{y^2+1}}dy=K$$ $$y =\sinh(z)$$ then $$k=\int\frac{1}{\cosh(z)}\cosh(z)dz = z+C= \sinh^{-1}(y)+C$$ so that $$I=\frac{7}{8}y\sqrt{y^2+1} +\frac{7}{8}\sinh^{-1}(y)$$ $$ y =\frac{2}{\sqrt{7}}(\frac{1}{2}+x)$$ so $$I=\frac{\sqrt{7}}{4}(\frac{1}{2}+x)\sqrt{(\frac{2}{\sqrt{7}}(\frac{1}{2}+x))^2+1} +\frac{7}{8}\sinh^{-1}(\frac{2}{\sqrt{7}}(\frac{1}{2}+x))+C$$