Найти интегралы:
\(1) \int \frac{x-2}{\sqrt{x^3}} dx \)
\( 2) \int \frac{(2\sqrt{x}+1)^2}{x^2} dx \)

Решение

\( 1) \int \frac{x-2}{\sqrt{x^3}} dx = \int \frac{x-2}{x-\sqrt{x}}dx = \int \frac{1}{x^{\frac{1}{2}}} - \frac{2}{x^{\frac{3}{2}}} dx = 2x^\frac{1}{2} - 2 \frac{1}{ (\frac{3}{2}-1)x^{\frac{3}{2}-1} } + C =\)
\(= 2x^\frac{1}{2}- 2\frac{2}{x^\frac{1}{2}}+ C = 2\sqrt{x} - 4\frac{1}{\sqrt{x}} + C = \frac{2x-4}{\sqrt{x}}+C = \frac{2(x-2)}{\sqrt{x}} + C\)

\(2) \int \frac{(2\sqrt{x}+1)^2}{x^2} dx = \int \frac{(x\sqrt{x}+1)^2}{x^2} dx = \int \frac{4x+4\sqrt{x} +1}{x^2} dx = \int \frac{4}{x} + \frac{4}{x\sqrt{x}} +\frac{1}{x^2} dx \)=
\(4\ln |x| - 4\frac{1}{(\frac{3}{2}-1 )x^{\frac{3}{2}-1}} - \frac{1}{(2-1)x} +C = 4\ln|x| - \frac{2*4}{x^\frac{1}{2}} - \frac{1}{x}+C = 4\ln |x| - \frac{8}{\sqrt{x}} - \frac{1}{x} +C \)

Ответ: \( 1)\frac{2(x-2)}{\sqrt{x}} + C\) \( 2) 4\ln |x| - \frac{8}{\sqrt{x}} - \frac{1}{x} +C \).