What is the integral of (5+t+t^2)/(sqrt(t)) ?

The integral of (5+t+t^2)/(sqrt(t)) is 10sqrt(t)+2/3 t^{3/2}+2/5 t^{5/2}+C

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integral of (5+t+t^2)/(sqrt(t))

\int \frac{5 + t + t ^{2}}{t} d t

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

Solution steps

\int \frac{5 + t + t ^{2}}{t} d t

Expand

\frac{5 + t + t ^{2}}{t} : \frac{5}{t} + t + t^{\frac{3}{2}}

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

\int \frac{5}{t} d t = 10 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}}

= 10 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}}

Add a constant to the solution

= 10 t + \frac{2}{3} t^{\frac{3}{2}} + \frac{2}{5} t^{\frac{5}{2}} + C

x \frac{d y}{d x} = y - \frac{1}{y}

x \to - 5 - lim (\frac{∣ x + 5 ∣}{x + 5})

\int x^{5} e^{- x^{6}} d x

x + y (\frac{d y}{d x}) = 0

\int (x x + 6) d x

What is the integral of (5+t+t^2)/(sqrt(t)) ?
The integral of (5+t+t^2)/(sqrt(t)) is 10sqrt(t)+2/3 t^{3/2}+2/5 t^{5/2}+C