Integral dari $$$\frac{\ln^{2}\left(x\right)}{x^{2}}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{\ln^{2}\left(x\right)}{x^{2}}\, dx$$$.
Solusi
Misalkan $$$u=\frac{1}{x}$$$.
Kemudian $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$\frac{dx}{x^{2}} = - du$$$.
Dengan demikian,
$${\color{red}{\int{\frac{\ln{\left(x \right)}^{2}}{x^{2}} d x}}} = {\color{red}{\int{\left(- \ln{\left(u \right)}^{2}\right)d u}}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=-1$$$ dan $$$f{\left(u \right)} = \ln{\left(u \right)}^{2}$$$:
$${\color{red}{\int{\left(- \ln{\left(u \right)}^{2}\right)d u}}} = {\color{red}{\left(- \int{\ln{\left(u \right)}^{2} d u}\right)}}$$
Untuk integral $$$\int{\ln{\left(u \right)}^{2} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.
Misalkan $$$\operatorname{m}=\ln{\left(u \right)}^{2}$$$ dan $$$\operatorname{dv}=du$$$.
Maka $$$\operatorname{dm}=\left(\ln{\left(u \right)}^{2}\right)^{\prime }du=\frac{2 \ln{\left(u \right)}}{u} du$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{1 d u}=u$$$ (langkah-langkah dapat dilihat di »).
Oleh karena itu,
$$- {\color{red}{\int{\ln{\left(u \right)}^{2} d u}}}=- {\color{red}{\left(\ln{\left(u \right)}^{2} \cdot u-\int{u \cdot \frac{2 \ln{\left(u \right)}}{u} d u}\right)}}=- {\color{red}{\left(u \ln{\left(u \right)}^{2} - \int{2 \ln{\left(u \right)} d u}\right)}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ dengan $$$c=2$$$ dan $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:
$$- u \ln{\left(u \right)}^{2} + {\color{red}{\int{2 \ln{\left(u \right)} d u}}} = - u \ln{\left(u \right)}^{2} + {\color{red}{\left(2 \int{\ln{\left(u \right)} d u}\right)}}$$
Untuk integral $$$\int{\ln{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{m} \operatorname{dv} = \operatorname{m}\operatorname{v} - \int \operatorname{v} \operatorname{dm}$$$.
Misalkan $$$\operatorname{m}=\ln{\left(u \right)}$$$ dan $$$\operatorname{dv}=du$$$.
Maka $$$\operatorname{dm}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{1 d u}=u$$$ (langkah-langkah dapat dilihat di »).
Oleh karena itu,
$$- u \ln{\left(u \right)}^{2} + 2 {\color{red}{\int{\ln{\left(u \right)} d u}}}=- u \ln{\left(u \right)}^{2} + 2 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=- u \ln{\left(u \right)}^{2} + 2 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$
Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:
$$- u \ln{\left(u \right)}^{2} + 2 u \ln{\left(u \right)} - 2 {\color{red}{\int{1 d u}}} = - u \ln{\left(u \right)}^{2} + 2 u \ln{\left(u \right)} - 2 {\color{red}{u}}$$
Ingat bahwa $$$u=\frac{1}{x}$$$:
$$- 2 {\color{red}{u}} + 2 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} - {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}^{2} = - 2 {\color{red}{\frac{1}{x}}} + 2 {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)} - {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)}^{2}$$
Oleh karena itu,
$$\int{\frac{\ln{\left(x \right)}^{2}}{x^{2}} d x} = - \frac{\ln{\left(\frac{1}{x} \right)}^{2}}{x} + \frac{2 \ln{\left(\frac{1}{x} \right)}}{x} - \frac{2}{x}$$
Sederhanakan:
$$\int{\frac{\ln{\left(x \right)}^{2}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} - 2}{x}$$
Tambahkan konstanta integrasi:
$$\int{\frac{\ln{\left(x \right)}^{2}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{2} - 2 \ln{\left(x \right)} - 2}{x}+C$$
Jawaban
$$$\int \frac{\ln^{2}\left(x\right)}{x^{2}}\, dx = \frac{- \ln^{2}\left(x\right) - 2 \ln\left(x\right) - 2}{x} + C$$$A