5. int(x^2dx)/(sqrt(x^2)-2)=（2分） A. (x^2)/(2)sqrt(x^2)-2+ln|x+sqrt(x^2)-2|+c B. (x^2)/(2)sqrt(x^2)-2-ln|x+sqrt(x^2)-2|+c C. (x)/(2)sqrt(x^2)-2+ln|x+sqrt(x^2)-2|+c D. (x)/(2)sqrt(x^2)-2-ln|x+sqrt(x^2)-2|+c

为了求解积分 $\int \frac{x^2 \, dx}{\sqrt{x^2 - 2}}$，我们可以使用换元法。设 $x = \sqrt{2} \sec \theta$，则 $dx = \sqrt{2} \sec \theta \tan \theta \, d\theta$。代入积分，我们得到： \[ \int \frac{x^2 \, dx}{\sqrt{x^2 - 2}} = \int \frac{(\sqrt{2} \sec \theta)^2 \cdot \sqrt{2} \sec \theta \tan \theta \, d\theta}{\sqrt{(\sqrt{2} \sec \theta)^2 - 2}} = \int \frac{2 \sec^2 \theta \cdot \sqrt{2} \sec \theta \tan \theta \, d\theta}{\sqrt{2 \sec^2 \theta - 2}}. \] 由于 $\sec^2 \theta - 1 = \tan^2 \theta$，我们有： \[ \int \frac{2 \sec^2 \theta \cdot \sqrt{2} \sec \theta \tan \theta \, d\theta}{\sqrt{2 \tan^2 \theta}} = \int \frac{2 \sec^2 \theta \cdot \sqrt{2} \sec \theta \tan \theta \, d\theta}{\sqrt{2} \tan \theta} = \int 2 \sec^3 \theta \, d\theta. \] 现在，我们需要计算 $\int 2 \sec^3 \theta \, d\theta$。使用分部积分法，设 $u = \sec \theta$ 和 $dv = \sec^2 \theta \, d\theta$，则 $du = \sec \theta \tan \theta \, d\theta$ 和 $v = \tan \theta$。因此，我们有： \[ \int \sec^3 \theta \, d\theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta \, d\theta = \sec \theta \tan \theta - \int \sec \theta (\sec^2 \theta - 1) \, d\theta = \sec \theta \tan \theta - \int \sec^3 \theta \, d\theta + \int \sec \theta \, d\theta. \] 将 $\int \sec^3 \theta \, d\theta$ 移到等式的一边，我们得到： \[ 2 \int \sec^3 \theta \, d\theta = \sec \theta \tan \theta + \int \sec \theta \, d\theta = \sec \theta \tan \theta + \ln|\sec \theta + \tan \theta| + C. \] 因此， \[ \int \sec^3 \theta \, d\theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln|\sec \theta + \tan \theta| + C. \] 代回 $x = \sqrt{2} \sec \theta$，我们有 $\sec \theta = \frac{x}{\sqrt{2}}$ 和 $\tan \theta = \sqrt{\sec^2 \theta - 1} = \sqrt{\left(\frac{x}{\sqrt{2}}\right)^2 - 1} = \sqrt{\frac{x^2}{2} - 1} = \frac{\sqrt{x^2 - 2}}{\sqrt{2}}$。因此， \[ \int \sec^3 \theta \, d\theta = \frac{1}{2} \cdot \frac{x}{\sqrt{2}} \cdot \frac{\sqrt{x^2 - 2}}{\sqrt{2}} + \frac{1}{2} \ln \left| \frac{x}{\sqrt{2}} + \frac{\sqrt{x^2 - 2}}{\sqrt{2}} \right| + C = \frac{1}{2} \cdot \frac{x \sqrt{x^2 - 2}}{2} + \frac{1}{2} \ln \left| \frac{x + \sqrt{x^2 - 2}}{\sqrt{2}} \right| + C = \frac{x \sqrt{x^2 - 2}}{4} + \frac{1}{2} \ln |x + \sqrt{x^2 - 2}| - \frac{1}{2} \ln \sqrt{2} + C. \] 由于 $- \frac{1}{2} \ln \sqrt{2}$ 是一个常数，可以合并到 $C$ 中，我们得到： \[ \int \sec^3 \theta \, d\theta = \frac{x \sqrt{x^2 - 2}}{4} + \frac{1}{2} \ln |x + \sqrt{x^2 - 2}| + C. \] 因此， \[ \int \frac{x^2 \, dx}{\sqrt{x^2 - 2}} = 2 \int \sec^3 \theta \, d\theta = 2 \left( \frac{x \sqrt{x^2 - 2}}{4} + \frac{1}{2} \ln |x + \sqrt{x^2 - 2}| + C \right) = \frac{x \sqrt{x^2 - 2}}{2} + \ln |x + \sqrt{x^2 - 2}| + C. \] 所以，正确答案是 $\boxed{C}$。