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人気のある微分積分 >

integral from 0 to 4 of (8t)/((t-5)^2)

解

\int_{0}^{4} \frac{8 t}{( t - 5 ) ^{2}} d t

解

8 (- ln (5) + 4)

+1

十進法表記

19.12449 \dots

解答ステップ

\int_{0}^{4} \frac{8 t}{( t - 5 ) ^{2}} d t

定数を除く:

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

= 8 \cdot \int_{0}^{4} \frac{t}{( t - 5 ) ^{2}} d t

u置換積分法を適用する

= 8 \cdot \int_{- 5}^{- 1} \frac{u + 5}{u ^{2}} d u

拡張

\frac{u + 5}{u ^{2}} : \frac{1}{u} + \frac{5}{u ^{2}}

= 8 \cdot \int_{- 5}^{- 1} \frac{1}{u} + \frac{5}{u ^{2}} 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

= 8 (\int_{- 5}^{- 1} \frac{1}{u} d u + \int_{- 5}^{- 1} \frac{5}{u ^{2}} d u)

\int_{- 5}^{- 1} \frac{1}{u} d u = - ln (5)

\int_{- 5}^{- 1} \frac{5}{u ^{2}} d u = 4

= 8 (- ln (5) + 4)

グラフ

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