What is the integral of (A+BT+CT^2+DT^2) 1/T ?

The integral of (A+BT+CT^2+DT^2) 1/T is Aln|T|+BT+(CT^2)/2+(DT^2)/2+C

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

\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

Solution steps

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

Simplify

(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

Expand

\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

Apply the Sum Rule:

\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}

Add a constant to the solution

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

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What is the integral of (A+BT+CT^2+DT^2) 1/T ?
The integral of (A+BT+CT^2+DT^2) 1/T is Aln|T|+BT+(CT^2)/2+(DT^2)/2+C