proof verification - Verify integration of $intfrac{sqrt{2-x-x^2}}{x^2}dx$

This is exercise 6.25.40 from Tom Apostol's Calculus I. I would like to ask someone to verify my solution, the result I got differs from the one provided in the book.

Evaluate the following integral: $\int\frac{\sqrt{2-x-x^2}}{x^2}dx$

$x \in [{-2,0}) \cup ({0,1}]$

As suggested in the book, we multiply both numerator and denumerator by $\sqrt{2-x-x^2}$. This removes the endpoints from the integrand's domain, but the definite integral we calculate with the antiderivative of this new function will be still the proper integral between any two points of the original domain: we only remove a finite number of points from the original domain and the domain of the resulting antiderivative will be the same as the original domain.

$$I=I_1+I_2=\int\frac{2-x}{x^2\sqrt{2-x-x^2}}dx-\int\frac1{\sqrt{2-x-x^2}}dx \tag{1}$$

Evaluating first $I_1$ by substituting $t=\frac1{x} \; \text, \; dx=-\frac1{t^2}dt \text:$

$$I_1=-\frac1{\sqrt2}\int\frac{2t-1}{\frac{t}{|t|}\sqrt{\left(t-\frac14\right)^2-\left(\frac34\right)^2}}dt \tag{2}$$

Substituting again $\frac34\sec{u}=t-\frac14 \; \text, \; dt=\frac34\sec{u}\tan{u}du \; \text, \; t=\frac{3\sec{u}+1}{4} \; \text, \; u=\operatorname{arcsec}{\frac{4t-1}{3}}$, by considering the sign of $t$ and $\tan{u}$ in the integrand's two sub-domains:

a) $x \in (-2, 0): t<-\frac12 \; \text, \; \frac{4t-1}{3}<-1 \; \text, \; u \in(\frac\pi2,\pi) \; \text, \; \tan{u}<0$
b) $x \in (0, 1): t>1 \; \text, \; \frac{4t-1}{3}>1 \; \text, \; u \in(0,\frac\pi2) \; \text, \; \tan{u}>0$

$$I_1=-\frac{1}{2\sqrt2}\int3\sec^2{u}-\sec{u}=-\frac1{2\sqrt2}\left(3\tan{u}-\log\left|\tan{u}+\sec{u}\right|\right)+C_1 \tag{3}$$

$$\sec{u}=\sec\operatorname{arcsec}\frac{4t-1}{3}=\frac{4t-1}{3}=\frac{4-x}{3x} \tag{4}$$

$$\tan^2{u}=\tan^2\operatorname{arcsec}\frac{4t-1}{3}=\left(\frac{4t-1}{3}\right)^2-1=\frac89\frac{2-x-x^2}{x^2} \tag{5}$$

Considering again cases a) and b):

$$\tan{u}=\frac{2\sqrt2}{3}\frac{\sqrt{2-x-x^2}}{x} \tag{6}$$

$$I_1=-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{2\sqrt2}{3}\left(\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right)\right|+C_1= \tag{7}$$

$$-\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+C'_1 \tag{8}$$

Evaluating now $I_2$:

$$I_2=-\int\frac1{\sqrt{\left(\frac32\right)^2-\left(x+\frac12\right)^2}}dx \tag{9}$$

Substituting $\frac32\sin{z}=x+\frac12 \; \text, \; dx=\frac32\cos{z}dz \; \text, \; z=\operatorname{arcsin}{\frac{2x+1}{3}}$:

$$I_2 = -\int dz = -\operatorname{arcsin}{\frac{2x+1}{3}}+C_2 \tag{10}$$

The final result:

$$I = I_1 + I_2 = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|-\operatorname{arcsin}{\frac{2x+1}{3}}+C \tag{11}$$

The solution provided in the book:

$$I = -\frac{\sqrt{2-x-x^2}}{x}+\frac1{2\sqrt2}\log\left(\frac{\sqrt{2-x-x^2}}{x}-\frac1{2\sqrt2}\right)-\operatorname{arcsin}\frac{2x+1}{3}+C$$

Answer

The solution in the book is most certainly a typo, your proof seems fine to me. As a confirmation, Mathematica evaluates the integral to be:

$$I=-\dfrac {\sqrt {2 - x - x^2}} x + \dfrac1 {2\sqrt {2}}\left[\log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x| \right] \qquad - \arcsin\left (\dfrac {2 x + 1} {3} \right) + \rm C_1,$$ which is the same as your proposed solution since $$\begin{align} \log\left |4 - x + 2\sqrt {2}\sqrt {2 - x - x^2} \right| - \log |x|&=\log\left|\dfrac{{2\sqrt{2}\sqrt{2-x-x^2}}+{4-x}}{2\sqrt{2}x}\right|+{\rm C_2} \\ &=\log\left|\frac{\sqrt{2-x-x^2}+\sqrt2}{x}-\frac1{2\sqrt2}\right|+{\rm C_2},\end{align}$$ so they only differ by a constant.