Нужно решить методом первой подстановки Эйлера, пожалуйста.

$\displaystyle \int\frac{\sqrt{2x+x^2}}{x^2}dx=\int\frac{t^2+2t}{2(1+t)}*\frac{4(1+t)^2}{t^4}*\frac{t^2+2t}{2(1+t)^2}dt=\\=\int\frac{(t+2)^2}{t^2(1+t)}dt=4\int\frac{dt}{t^2}+\int\frac{dt}{t+1}=-\frac{4}{t}+ln|t+1|+C=\\=-\frac{4}{\sqrt{2x+x^2}+x}+ln|\sqrt{2x+x^2}+x+1|+C\\\\\\\sqrt{2x+x^2}+x=t\\2x+x^2=t^2-2tx+x^2\\2x+2tx=t^2\\x=\frac{t^2}{2(1+t)}\\x^2=\frac{t^4}{4(1+t)^2}\\dx=\frac{t^2+2t}{2(1+t^2)^2}dt\\\sqrt{2x+x^2}=t-\frac{t^2}{2(1+t)}=\frac{t^2+2t}{2(1+t)}$

Проверка:
$(-\frac{4}{\sqrt{2x+x^2}+x}+ln|\sqrt{2x+x^2}+x+1|+C)'=\\=\frac{4}{(\sqrt{2x+x^2}+x)^2}(\frac{x+1}{\sqrt{2x+x^2}}+1)+\frac{1}{\sqrt{2x+x^2}+x+1}(\frac{x+1}{\sqrt{2x+x^2}}+1)=\\=(\frac{x+1+\sqrt{2x+x^2}}{\sqrt{2x+x^2}})(\frac{2+x}{x(1+x+\sqrt{2x+x^2})})=\frac{2+x}{x\sqrt{2x+x^2}}=\frac{2x+x^2}{x^2\sqrt{2x+x^2}}=\frac{\sqrt{2x+x^2}}{x^2}\\\\\\(\sqrt{2x+x^2}+x)^2=2x+x^2+x^2+2x\sqrt{2x+x^2}=\\=2x(1+x+\sqrt{2x+x^2})$