integral of 1/((t+2)^2(t+1))

解

\int \frac{1}{( t + 2 ) ^{2} ( t + 1 )} d t

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

解答ステップ

\int \frac{1}{( t + 2 ) ^{2} ( t + 1 )} d t

以下の部分分数を得る:

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

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

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{1}{t + 2} d t - \int \frac{1}{( t + 2 ) ^{2}} d t + \int \frac{1}{t + 1} d t

\int \frac{1}{t + 2} d t = ln ∣ t + 2 ∣

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

\int \frac{1}{t + 1} d t = ln ∣ t + 1 ∣

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

簡素化

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

定数を解答に追加する

= - ln ∣ t + 2 ∣ + \frac{1}{t + 2} + ln ∣ t + 1 ∣ + C