$$$\frac{1}{- \sqrt{3} x + \sqrt{2} x}$$$의 적분

$$$\int \frac{1}{- \sqrt{3} x + \sqrt{2} x}\, dx$$$을(를) 구하시오.

$$$u=- \sqrt{3} x + \sqrt{2} x$$$라 하자.

그러면 $$$du=\left(- \sqrt{3} x + \sqrt{2} x\right)^{\prime }dx = \left(- \sqrt{3} + \sqrt{2}\right) dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = \frac{du}{- \sqrt{3} + \sqrt{2}}$$$임을 얻습니다.

따라서,

$${\color{red}{\int{\frac{1}{- \sqrt{3} x + \sqrt{2} x} d x}}} = {\color{red}{\int{\frac{1}{u \left(- \sqrt{3} + \sqrt{2}\right)} d u}}}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\frac{1}{- \sqrt{3} + \sqrt{2}}$$$와 $$$f{\left(u \right)} = \frac{1}{u}$$$에 적용하세요:

$${\color{red}{\int{\frac{1}{u \left(- \sqrt{3} + \sqrt{2}\right)} d u}}} = {\color{red}{\frac{\int{\frac{1}{u} d u}}{- \sqrt{3} + \sqrt{2}}}}$$

$$$\frac{1}{u}$$$의 적분은 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$\frac{{\color{red}{\int{\frac{1}{u} d u}}}}{- \sqrt{3} + \sqrt{2}} = \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{- \sqrt{3} + \sqrt{2}}$$

다음 $$$u=- \sqrt{3} x + \sqrt{2} x$$$을 기억하라:

$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{- \sqrt{3} + \sqrt{2}} = \frac{\ln{\left(\left|{{\color{red}{\left(- \sqrt{3} x + \sqrt{2} x\right)}}}\right| \right)}}{- \sqrt{3} + \sqrt{2}}$$

따라서,

$$\int{\frac{1}{- \sqrt{3} x + \sqrt{2} x} d x} = \frac{\ln{\left(\left|{- \sqrt{3} x + \sqrt{2} x}\right| \right)}}{- \sqrt{3} + \sqrt{2}}$$

간단히 하시오:

$$\int{\frac{1}{- \sqrt{3} x + \sqrt{2} x} d x} = \frac{\ln{\left(\left|{x}\right| \right)} + \ln{\left(- \sqrt{2} + \sqrt{3} \right)}}{- \sqrt{3} + \sqrt{2}}$$

적분 상수를 추가하세요:

$$\int{\frac{1}{- \sqrt{3} x + \sqrt{2} x} d x} = \frac{\ln{\left(\left|{x}\right| \right)} + \ln{\left(- \sqrt{2} + \sqrt{3} \right)}}{- \sqrt{3} + \sqrt{2}}+C$$

$$$\int \frac{1}{- \sqrt{3} x + \sqrt{2} x}\, dx = \frac{\ln\left(\left|{x}\right|\right) + \ln\left(- \sqrt{2} + \sqrt{3}\right)}{- \sqrt{3} + \sqrt{2}} + C$$$A