integral from 1 to t of 2/(x^2+4x+3)

解

\int_{1}^{t} \frac{2}{x ^{2} + 4 x + 3} d x

ln (2) + ln ∣ t + 1 ∣ - ln ∣ t + 3 ∣

解答ステップ

\int_{1}^{t} \frac{2}{x ^{2} + 4 x + 3} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int_{1}^{t} \frac{1}{3 + x ^{2} + 4 x} d x

以下の部分分数を得る:

\frac{1}{3 + x ^{2} + 4 x} : \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x + 3 )}

= 2 \cdot \int_{1}^{t} \frac{1}{2 ( x + 1 )} - \frac{1}{2 ( x + 3 )} 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

= 2 (\int_{1}^{t} \frac{1}{2 ( x + 1 )} d x - \int_{1}^{t} \frac{1}{2 ( x + 3 )} d x)

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

\int_{1}^{t} \frac{1}{2 ( x + 3 )} d x = \frac{1}{2} ln ∣ t + 3 ∣ - ln (2)

= 2 (\frac{1}{2} (ln ∣ t + 1 ∣ - ln (2)) - (\frac{1}{2} ln ∣ t + 3 ∣ - ln (2)))

簡素化

2 (\frac{1}{2} (ln ∣ t + 1 ∣ - ln (2)) - (\frac{1}{2} ln ∣ t + 3 ∣ - ln (2))) : ln (2) + ln ∣ t + 1 ∣ - ln ∣ t + 3 ∣

= ln (2) + ln ∣ t + 1 ∣ - ln ∣ t + 3 ∣