integral from x to 2x of (at)/(bt^2+c)

解

\int_{x}^{2 x} \frac{a t}{b t ^{2} + c} d t

\frac{a ( ln c + 4 x ^{2} b - ln c + x ^{2} b )}{2 b}

解答ステップ

\int_{x}^{2 x} \frac{a t}{b t ^{2} + c} d t

定数を除く:

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

= a \cdot \int_{x}^{2 x} \frac{t}{c + b t ^{2}} d t

u置換積分法を適用する

= a \cdot \int_{c + b x^{2}}^{c + 4 x^{2} b} \frac{1}{2 b u} d u

定数を除く:

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

= a \frac{1}{2 b} \cdot \int_{c + b x^{2}}^{c + 4 x^{2} b} \frac{1}{u} d u

共通積分を使用する:

\int \frac{1}{u} d u = ln (∣ u ∣)

= a \frac{1}{2 b} [ln ∣ u ∣]_{c + b x^{2}}^{c + 4 x^{2} b}

限度を計算する:

ln c + 4 x^{2} b - ln c + b x^{2}

= a \frac{1}{2 b} (ln c + 4 x^{2} b - ln c + b x^{2})

簡素化

= \frac{a ( ln c + 4 x ^{2} b - ln c + x ^{2} b )}{2 b}