integral of (A+BT+CT^2+DT^2) 1/T

解

\int (A + BT + C T^{2} + D T^{2}) \frac{1}{T} d T

A ln ∣ T ∣ + BT + \frac{C T ^{2}}{2} + \frac{D T ^{2}}{2} + C

解答ステップ

\int (A + BT + C T^{2} + D T^{2}) \frac{1}{T} d T

簡素化

(A + BT + C T^{2} + D T^{2}) \frac{1}{T} : \frac{A + BT + C T ^{2} + D T ^{2}}{T}

= \int \frac{A + BT + C T ^{2} + D T ^{2}}{T} d T

拡張

\frac{A + BT + C T ^{2} + D T ^{2}}{T} : \frac{A}{T} + B + CT + D T

= \int \frac{A}{T} + B + CT + D T d T

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 \frac{A}{T} d T + \int B d T + \int CT d T + \int D T d T

\int \frac{A}{T} d T = A ln ∣ T ∣

\int B d T = BT

\int CT d T = \frac{C T ^{2}}{2}

\int D T d T = \frac{D T ^{2}}{2}

= A ln ∣ T ∣ + BT + \frac{C T ^{2}}{2} + \frac{D T ^{2}}{2}

定数を解答に追加する

= A ln ∣ T ∣ + BT + \frac{C T ^{2}}{2} + \frac{D T ^{2}}{2} + C