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

The integral of (2t-4+6sqrt(t))/(sqrt(t)) is 4/3 t^{3/2}-8sqrt(t)+6t+C

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integral of (2t-4+6sqrt(t))/(sqrt(t))

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

\frac{4}{3} t^{\frac{3}{2}} - 8 t + 6 t + C

Solution steps

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

Expand

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

= \int 2 t - \frac{4}{t} + 6 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 t d t - \int \frac{4}{t} d t + \int 6 d t

\int 2 t d t = \frac{4}{3} t^{\frac{3}{2}}

\int \frac{4}{t} d t = 8 t

\int 6 d t = 6 t

= \frac{4}{3} t^{\frac{3}{2}} - 8 t + 6 t

Add a constant to the solution

= \frac{4}{3} t^{\frac{3}{2}} - 8 t + 6 t + C

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What is the integral of (2t-4+6sqrt(t))/(sqrt(t)) ?
The integral of (2t-4+6sqrt(t))/(sqrt(t)) is 4/3 t^{3/2}-8sqrt(t)+6t+C