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

解

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

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

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

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

簡素化

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

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

定数を解答に追加する

= \frac{1}{2} arctan (t) - \frac{t}{2 ( 1 + t ^{2} )} + C