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

解答

\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

求解步骤

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

化简

(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

乘开

\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

使用积分加法定则:

\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}

解答补常数

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

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