Question

помагите решить задачу интегралы.​

Answer 1

Ответ:

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Answer 2

$\int \dfrac{dx}{\sqrt{x}-1}=\Big[\ x=t^2\ ,\ t=\sqrt{x}\ ,\ dx=2t\, dt\ \Big]=\int \dfrac{2t\, dt}{t-1}=2\int \Big(1+\dfrac{1}{t-1}\Big)\, dt=\\\\\\=2\cdot t+2\cdot ln|t-1|+C=2\sqrt{x}+2\cdot ln|\sqrt{x}-1|+C\ ;\\\\\\\\\int \limits _4^9\dfrac{dx}{\sqrt{x}-1}=\Big(2\sqrt{x}+2\, ln|\sqrt{x}-1|\Big)\Big|_4^9=\Big(2\cdot 3+2\, ln|2|\Big)-\Big(2\cdot 2+2\, ln|1|\Big)=\\\\\\=6+2\, ln2-4-0=2+2\, ln2=2+ln4$