What is the integral of sqrt(t)-1/(sqrt(t)) ?

The integral of sqrt(t)-1/(sqrt(t)) is 2/3 t^{3/2}-2sqrt(t)+C

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integral of sqrt(t)-1/(sqrt(t))

\int t - \frac{1}{t} d t

\frac{2}{3} t^{\frac{3}{2}} - 2 t + C

Solution steps

\int t - \frac{1}{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 t d t - \int \frac{1}{t} d t

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

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

= \frac{2}{3} t^{\frac{3}{2}} - 2 t

Add a constant to the solution

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

(1, 5), (- 2, - 4)

x y^{^{'}} + y = 8 x y^{2}, y (1) = 5

\int_{0}^{\frac{π}{2}} 9 sin^{2} (2 x) d x

\frac{d y}{d x} = - \frac{( 2 x y + y ^{2} + 1 )}{x ^{2} + 2 x y}

\frac{1 x ^{3}}{2 x ^{4}}

What is the integral of sqrt(t)-1/(sqrt(t)) ?
The integral of sqrt(t)-1/(sqrt(t)) is 2/3 t^{3/2}-2sqrt(t)+C