What is the integral of 2sqrt(t)-t-9/(t^2) ?

The integral of 2sqrt(t)-t-9/(t^2) is 4/3 t^{3/2}-(t^2)/2+9/t+C

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integral of 2sqrt(t)-t-9/(t^2)

\int 2 t - t - \frac{9}{t ^{2}} d t

\frac{4}{3} t^{\frac{3}{2}} - \frac{t ^{2}}{2} + \frac{9}{t} + C

Solution steps

\int 2 t - t - \frac{9}{t ^{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 2 t d t - \int t d t - \int \frac{9}{t ^{2}} d t

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

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

\int \frac{9}{t ^{2}} d t = - \frac{9}{t}

= \frac{4}{3} t^{\frac{3}{2}} - \frac{t ^{2}}{2} - (- \frac{9}{t})

Simplify

= \frac{4}{3} t^{\frac{3}{2}} - \frac{t ^{2}}{2} + \frac{9}{t}

Add a constant to the solution

= \frac{4}{3} t^{\frac{3}{2}} - \frac{t ^{2}}{2} + \frac{9}{t} + C

x \to 0 lim (x^{2} - 4)

\int \frac{x ^{2}}{( 1 - y ^{2} )}

e^{a x}

\int (\frac{4}{x} - \frac{16}{x ^{2}}) d x

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{9}})

What is the integral of 2sqrt(t)-t-9/(t^2) ?
The integral of 2sqrt(t)-t-9/(t^2) is 4/3 t^{3/2}-(t^2)/2+9/t+C