What is the integral of (4t+6)/(2t+1) ?

The integral of (4t+6)/(2t+1) is 2t+2ln|2t+1|+C

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integral of (4t+6)/(2t+1)

\int \frac{4 t + 6}{2 t + 1} d t

2 t + 2 ln ∣ 2 t + 1 ∣ + C

Solution steps

\int \frac{4 t + 6}{2 t + 1} d t

Long division

\frac{4 t + 6}{2 t + 1} : 2 + \frac{4}{2 t + 1}

= \int 2 + \frac{4}{2 t + 1} 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 2 d t + \int \frac{4}{2 t + 1} d t

\int 2 d t = 2 t

\int \frac{4}{2 t + 1} d t = 2 ln ∣ 2 t + 1 ∣

= 2 t + 2 ln ∣ 2 t + 1 ∣

Add a constant to the solution

= 2 t + 2 ln ∣ 2 t + 1 ∣ + C

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What is the integral of (4t+6)/(2t+1) ?
The integral of (4t+6)/(2t+1) is 2t+2ln|2t+1|+C