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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

定数を除く:

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

= 2 \cdot \int \frac{t ^{2}}{2 t ^{2} + 1} d t

置換積分法を適用する

= 2 \cdot \int \frac{u ^{2}}{2 2 ( u ^{2} + 1 )} d u

定数を除く:

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

= 2 \cdot \frac{1}{2 2} \cdot \int \frac{u ^{2}}{u ^{2} + 1} d u

\frac{u ^{2}}{u ^{2} + 1} = - \frac{1}{u ^{2} + 1} + 1

= 2 \cdot \frac{1}{2 2} \cdot \int - \frac{1}{u ^{2} + 1} + 1 d u

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

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

\int 1 d u = u

= 2 \cdot \frac{1}{2 2} (- arctan (u) + u)

代用を戻す

u = 2 t

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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