integral from 0 to 6x^2 of (t/(t+36))

解

\int_{0}^{6 x^{2}} (\frac{t}{t + 36}) d t

6 x^{2} - 36 (ln 6 x^{2} + 36 - 2 ln (6))

解答ステップ

\int_{0}^{6 x^{2}} \frac{t}{t + 36} d t

u置換積分法を適用する

= \int_{36}^{6 x^{2} + 36} \frac{u - 36}{u} d u

拡張

\frac{u - 36}{u} : 1 - \frac{36}{u}

= \int_{36}^{6 x^{2} + 36} 1 - \frac{36}{u} 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

= \int_{36}^{6 x^{2} + 36} 1 d u - \int_{36}^{6 x^{2} + 36} \frac{36}{u} d u

\int_{36}^{6 x^{2} + 36} 1 d u = 6 x^{2}

\int_{36}^{6 x^{2} + 36} \frac{36}{u} d u = 36 (ln 6 x^{2} + 36 - 2 ln (6))

= 6 x^{2} - 36 (ln 6 x^{2} + 36 - 2 ln (6))