integral from 1 to x^3 of (g(t))/(4+t^2)

解

\int_{1}^{x^{3}} \frac{g ( t )}{4 + t ^{2}} d t

\frac{g ( ln 4 + x ^{6} - ln ( 5 ) )}{2}

解答ステップ

\int_{1}^{x^{3}} \frac{g t}{4 + t ^{2}} d t

定数を除く:

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

= g \cdot \int_{1}^{x^{3}} \frac{t}{4 + t ^{2}} d t

u置換積分法を適用する

= g \cdot \int_{5}^{4 + x^{6}} \frac{1}{2 u} d u

定数を除く:

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

= g \frac{1}{2} \cdot \int_{5}^{4 + x^{6}} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= g \frac{1}{2} [ln ∣ u ∣]_{5}^{4 + x^{6}}

限度を計算する:

ln 4 + x^{6} - ln (5)

= g \frac{1}{2} (ln 4 + x^{6} - ln (5))

簡素化

= \frac{g ( ln 4 + x ^{6} - ln ( 5 ) )}{2}