integral from 1 to e of ((1+Inx)^2)/x

解

\int_{1}^{e} \frac{( 1 + I n x ) ^{2}}{x} d x

1 + 2 I n e - 2 I n + \frac{( e ^{2} - 1 ) I ^{2} n ^{2}}{2}

解答ステップ

\int_{1}^{e} \frac{( 1 + I n x ) ^{2}}{x} d x

拡張

\frac{( 1 + I n x ) ^{2}}{x} : \frac{1}{x} + 2 I n + I^{2} n^{2} x

= \int_{1}^{e} \frac{1}{x} + 2 I n + I^{2} n^{2} x d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{1}^{e} \frac{1}{x} d x + \int_{1}^{e} 2 I n d x + \int_{1}^{e} I^{2} n^{2} x d x

\int_{1}^{e} \frac{1}{x} d x = 1

\int_{1}^{e} 2 I n d x = 2 I n e - 2 I n

\int_{1}^{e} I^{2} n^{2} x d x = \frac{( e ^{2} - 1 ) I ^{2} n ^{2}}{2}

= 1 + 2 I n e - 2 I n + \frac{( e ^{2} - 1 ) I ^{2} n ^{2}}{2}