簡約 1/2 ln|2x+1|+2ln|x-1|+C

解

簡約

\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

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

解答ステップ

\frac{1}{2} ln ∣ 2 x + 1 ∣ + 2 ln ∣ x - 1 ∣ + C

対数の規則を適用する:

a lo g_{c} (b) = lo g_{c} (b^{a})

\frac{1}{2} ln ∣ 2 x + 1 ∣ = ln (∣ 2 x + 1 ∣^{\frac{1}{2}})

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

対数の規則を適用する:

a lo g_{c} (b) = lo g_{c} (b^{a})

2 ln ∣ x - 1 ∣ = ln (∣ x - 1 ∣^{2})

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

対数の規則を適用する:

lo g_{c} (a) + lo g_{c} (b) = lo g_{c} (ab)

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

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

∣ x - 1 ∣^{2} = (x - 1)^{2}

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