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Ответ:

Применяем тригонометрическую подстановку .

[tex]\displaystyle \int\limits_1^{\sqrt8}\, \frac{\sqrt{(1+x^2)^3}}{x^2}\, dx=\Big[\ x=tg\, t\ ,\ dx=\frac{dt}{cos^2t}\ ,\ t=arctgx\ \Big]=\\\\\\=\int\limits_{\pi /4}^{arctg\sqrt8}\, \frac{\sqrt{(1+tg^2t)^3}}{tg^2t}\cdot \frac{dt}{cos^2t}=\int\limits_{\pi /4}^{arctg\sqrt8}\, \frac{\sqrt{\Big(\dfrac{1}{cos^2t}\Big)^3}}{tg^2t}\cdot \frac{dt}{cos^2t}=[/tex]

[tex]\displaystyle =\int\limits_{\pi /4}^{arctg\sqrt8}\, \frac{\dfrac{1}{cos^3t}}{tg^2t}\cdot \frac{dt}{cos^2t}=\int\limits_{\pi /4}^{arctg\sqrt8}\, \frac{1}{cos^3t\cdot \dfrac{sin^2t}{cos^2t}\cdot cos^2t}\, dt=\\\\\\=\int\limits_{\pi /4}^{arctg\sqrt8}\, \frac{dt}{sin^2t\cdot cos^3t}=I[/tex]

Найдём первообразную для последнего интеграла.

[tex]\displaystyle \int \frac{dt}{sin^2t\cdot cos^3t}=\int \frac{sin^2t+cos^2t}{sin^2t\cdot cos^3t}\, dt=\int \frac{sin^2t}{sin^2t\cdot cos^3t}\, dt+\int \frac{cos^2t}{sin^2t\cdot cos^3t}\, dt=\\\\\\=\int \frac{dt}{cos^3t}+\int \frac{dt}{sin^2t\cdot cost}=\int \frac{sin^2t+cos^2t}{cos^3t}\, dt+\int \frac{sin^2t+cos^2t}{sin^2t\cdot cost}\, dt=\\\\\\=\int \frac{sint\cdot sint\, dt}{cos^3t}+\int \frac{dt}{cost}+\int \frac{dt}{cost}+\int \frac{cost\, dt}{sin^2t}=(*)[/tex]

 Применим интегрирование по частям .

[tex]\displaystyle \star \ \int \frac{sint\cdot sint\, dt}{cos^3t}=\Big[u=sint\ ,\ du=cost\, dt\ ,\ dv=\frac{sint\, dt}{cos^3t}=-\int \frac{d(cost)}{cos^3t}=\\\\\\=\frac{1}{2cos^2t}\ \Big]=\frac{sint}{2cos^2t}-\int \frac{cost\, dt}{2cos^2t}=\frac{sint}{2cos^2t}-\frac{1}{2}\int \frac{dt}{cost}=\\\\\\=\frac{sint}{2cos^2t}-\frac{1}{2}ln\Big|\, tg(\frac{t}{2}+\frac{\pi}{4})\, \Big|+C\ ;[/tex]

[tex]\displaystyle (*)=\frac{sint}{2cos^2t}-\frac{1}{2}ln\Big|\, tg(\frac{t}{2}+\frac{\pi}{4})\, \Big|+2\int \frac{dt}{cost}+\int \frac{d(sint)}{sin^2t}=\\\\\\=\frac{sint}{2cos^2t}+\frac{3}{2}ln\Big|\, tg(\frac{t}{2}+\frac{\pi}{4})\, \Big|-\frac{1}{sint}+C\ ;[/tex]

[tex]\displaystyle I=\Big(\frac{sint}{2cos^2t}+\frac{3}{2}ln\Big|\, tg(\frac{t}{2}+\frac{\pi}{4})\, \Big|-\frac{1}{sint}\Big)\Big|_{\pi/4}^{arctg\sqrt8}=\\\\\\=\Big(\frac{\sqrt8}{3\cdot 2\cdot \frac{1}{9}}+\frac{3}{2}\, ln\Big|tg\Big(\frac{arctg\sqrt8}{2}+\frac{\pi}{4}\Big)\Big|-\frac{3}{\sqrt8}\Big)-\Big(\frac{1}{\sqrt2}+\frac{3}{2}\, ln\Big|tg\frac{3\pi }{8}\Big|-\sqrt2\Big)=[/tex]

  [tex]\displaystyle =\Big(\frac{3\sqrt8}{2}+\frac{3}{4}\, ln\Big|\, \frac{3+\sqrt8}{3-\sqrt8}\, \Big|-\frac{3}{\sqrt8}\Big)-\Big(\frac{1}{\sqrt2}+\frac{3}{4}\, ln\Big|\, \frac{\sqrt2+1}{\sqrt2-1}\, \Big|-\sqrt2\Big)[/tex]