Integral dari $$$e - \ln\left(x + 1\right)$$$

Temukan $$$\int \left(e - \ln\left(x + 1\right)\right)\, dx$$$.

Integralkan suku demi suku:

$${\color{red}{\int{\left(e - \ln{\left(x + 1 \right)}\right)d x}}} = {\color{red}{\left(\int{e d x} - \int{\ln{\left(x + 1 \right)} d x}\right)}}$$

Terapkan aturan konstanta $$$\int c\, dx = c x$$$ dengan $$$c=e$$$:

$$- \int{\ln{\left(x + 1 \right)} d x} + {\color{red}{\int{e d x}}} = - \int{\ln{\left(x + 1 \right)} d x} + {\color{red}{e x}}$$

Misalkan $$$u=x + 1$$$.

Kemudian $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (langkah-langkah dapat dilihat di »), dan kita memperoleh $$$dx = du$$$.

Integralnya menjadi

$$e x - {\color{red}{\int{\ln{\left(x + 1 \right)} d x}}} = e x - {\color{red}{\int{\ln{\left(u \right)} d u}}}$$

Untuk integral $$$\int{\ln{\left(u \right)} d u}$$$, gunakan integrasi parsial $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$.

Misalkan $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ dan $$$\operatorname{dv}=du$$$.

Maka $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$ (langkah-langkah dapat dilihat di ») dan $$$\operatorname{v}=\int{1 d u}=u$$$ (langkah-langkah dapat dilihat di »).

Jadi,

$$e x - {\color{red}{\int{\ln{\left(u \right)} d u}}}=e x - {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=e x - {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$

Terapkan aturan konstanta $$$\int c\, du = c u$$$ dengan $$$c=1$$$:

$$- u \ln{\left(u \right)} + e x + {\color{red}{\int{1 d u}}} = - u \ln{\left(u \right)} + e x + {\color{red}{u}}$$

Ingat bahwa $$$u=x + 1$$$:

$$e x + {\color{red}{u}} - {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} = e x + {\color{red}{\left(x + 1\right)}} - {\color{red}{\left(x + 1\right)}} \ln{\left({\color{red}{\left(x + 1\right)}} \right)}$$

Oleh karena itu,

$$\int{\left(e - \ln{\left(x + 1 \right)}\right)d x} = x + e x - \left(x + 1\right) \ln{\left(x + 1 \right)} + 1$$

Tambahkan konstanta integrasi (dan hapus konstanta dari ekspresi):

$$\int{\left(e - \ln{\left(x + 1 \right)}\right)d x} = x + e x - \left(x + 1\right) \ln{\left(x + 1 \right)}+C$$

$$$\int \left(e - \ln\left(x + 1\right)\right)\, dx = \left(x + e x - \left(x + 1\right) \ln\left(x + 1\right)\right) + C$$$A