integral of (2u-1)/(4u^2-1)

解

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

\frac{1}{4} ln 4 u^{2} - 1 + \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣) + C

解答ステップ

\int \frac{2 u - 1}{4 u ^{2} - 1} d u

拡張

\frac{2 u - 1}{4 u ^{2} - 1} : \frac{2 u}{4 u ^{2} - 1} - \frac{1}{4 u ^{2} - 1}

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{2 u}{4 u ^{2} - 1} d u - \int \frac{1}{4 u ^{2} - 1} d u

\int \frac{2 u}{4 u ^{2} - 1} d u = \frac{1}{4} ln 4 u^{2} - 1

\int \frac{1}{4 u ^{2} - 1} d u = - \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

= \frac{1}{4} ln 4 u^{2} - 1 - (- \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣))

簡素化

= \frac{1}{4} ln 4 u^{2} - 1 + \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣)

定数を解答に追加する

= \frac{1}{4} ln 4 u^{2} - 1 + \frac{1}{4} (ln ∣ 2 u + 1 ∣ - ln ∣ 2 u - 1 ∣) + C