integral of ((ln(x)-1)^2)/x

解

\int \frac{( ln ( x ) - 1 ) ^{2}}{x} d x

\frac{ln ^{3} ( x )}{3} - ln^{2} (x) + ln ∣ x ∣ + C

解答ステップ

\int \frac{( ln ( x ) - 1 ) ^{2}}{x} d x

拡張

\frac{( ln ( x ) - 1 ) ^{2}}{x} : \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} + \frac{1}{x}

= \int \frac{ln ^{2} ( x )}{x} - \frac{2 ln ( x )}{x} + \frac{1}{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 \frac{ln ^{2} ( x )}{x} d x - \int \frac{2 ln ( x )}{x} d x + \int \frac{1}{x} d x

\int \frac{ln ^{2} ( x )}{x} d x = \frac{ln ^{3} ( x )}{3}

\int \frac{2 ln ( x )}{x} d x = ln^{2} (x)

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{ln ^{3} ( x )}{3} - ln^{2} (x) + ln ∣ x ∣

定数を解答に追加する

= \frac{ln ^{3} ( x )}{3} - ln^{2} (x) + ln ∣ x ∣ + C