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

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

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

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

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

Solution steps

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

Expand

\frac{t - 2 t ^{4}}{t} : \frac{t}{t} - \frac{2 t ^{4}}{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{t}{t} d t - \int \frac{2 t ^{4}}{t} d t

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

\int \frac{2 t ^{4}}{t} d t = \frac{4}{9} t^{\frac{9}{2}}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x)^{2} + (y)^{4})

\frac{x ^{3}}{e ^{\frac{1}{x}}}

\frac{\partial}{\partial z} (\frac{x y ^{2}}{z ^{3}})

\frac{\partial}{\partial x} (\frac{y - 5 x}{y ^{2} + 4 x ^{2}})

\frac{\partial}{\partial x} (x^{0.5} y)

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