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

The integral of (4+t+t^2)/(sqrt(t)) is 8sqrt(t)+2/3 t^{3/2}+2/5 t^{5/2}+C

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integral of (4+t+t^2)/(sqrt(t))

\int \frac{4 + t + t ^{2}}{t} d t

8 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

Solution steps

\int \frac{4 + t + t ^{2}}{t} d t

Expand

\frac{4 + t + t ^{2}}{t} : \frac{4}{t} + t + t^{\frac{3}{2}}

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

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

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

\int t^{\frac{3}{2}} d t = \frac{2}{5} t^{\frac{5}{2}}

= 8 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}}

Add a constant to the solution

= 8 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

y = \frac{1}{x - 2}, at x = 6

\int e^{5} d x

\frac{d}{d x} ((2 x^{3} - 4 x^{2}) \cdot (3 x^{5} + x^{2}))

x \to i lim (\frac{z + i}{z - i})

f (x) = - 2 x^{2} + 8

What is the integral of (4+t+t^2)/(sqrt(t)) ?
The integral of (4+t+t^2)/(sqrt(t)) is 8sqrt(t)+2/3 t^{3/2}+2/5 t^{5/2}+C