Integral dari $$$\frac{x^{2} + 2 x + 1}{x^{2}}$$$
Kalkulator terkait: Kalkulator Integral Tentu dan Tak Wajar
Masukan Anda
Temukan $$$\int \frac{x^{2} + 2 x + 1}{x^{2}}\, dx$$$.
Solusi
Expand the expression:
$${\color{red}{\int{\frac{x^{2} + 2 x + 1}{x^{2}} d x}}} = {\color{red}{\int{\left(1 + \frac{2}{x} + \frac{1}{x^{2}}\right)d x}}}$$
Integralkan suku demi suku:
$${\color{red}{\int{\left(1 + \frac{2}{x} + \frac{1}{x^{2}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{2}} d x} + \int{\frac{2}{x} d x}\right)}}$$
Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=1$$$:
$$\int{\frac{1}{x^{2}} d x} + \int{\frac{2}{x} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{2}} d x} + \int{\frac{2}{x} d x} + {\color{red}{x}}$$
Terapkan aturan pangkat $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ dengan $$$n=-2$$$:
$$x + \int{\frac{2}{x} d x} + {\color{red}{\int{\frac{1}{x^{2}} d x}}}=x + \int{\frac{2}{x} d x} + {\color{red}{\int{x^{-2} d x}}}=x + \int{\frac{2}{x} d x} + {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=x + \int{\frac{2}{x} d x} + {\color{red}{\left(- x^{-1}\right)}}=x + \int{\frac{2}{x} d x} + {\color{red}{\left(- \frac{1}{x}\right)}}$$
Terapkan aturan pengali konstanta $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ dengan $$$c=2$$$ dan $$$f{\left(x \right)} = \frac{1}{x}$$$:
$$x + {\color{red}{\int{\frac{2}{x} d x}}} - \frac{1}{x} = x + {\color{red}{\left(2 \int{\frac{1}{x} d x}\right)}} - \frac{1}{x}$$
Integral dari $$$\frac{1}{x}$$$ adalah $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:
$$x + 2 {\color{red}{\int{\frac{1}{x} d x}}} - \frac{1}{x} = x + 2 {\color{red}{\ln{\left(\left|{x}\right| \right)}}} - \frac{1}{x}$$
Oleh karena itu,
$$\int{\frac{x^{2} + 2 x + 1}{x^{2}} d x} = x + 2 \ln{\left(\left|{x}\right| \right)} - \frac{1}{x}$$
Tambahkan konstanta integrasi:
$$\int{\frac{x^{2} + 2 x + 1}{x^{2}} d x} = x + 2 \ln{\left(\left|{x}\right| \right)} - \frac{1}{x}+C$$
Jawaban
$$$\int \frac{x^{2} + 2 x + 1}{x^{2}}\, dx = \left(x + 2 \ln\left(\left|{x}\right|\right) - \frac{1}{x}\right) + C$$$A