What is the integral of t(2t-3sqrt(t)) ?

The integral of t(2t-3sqrt(t)) is 2(1/5 t^{5/2}(2sqrt(t)-3)-(t^3)/(15))+C

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integral of t(2t-3sqrt(t))

\int t (2 t - 3 t) d t

2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15}) + C

Solution steps

\int t (2 t - 3 t) d t

Apply u-substitution

: \int 2 u^{3} (2 u^{2} - 3 u) d u

= \int 2 u^{3} (2 u^{2} - 3 u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int u^{3} (2 u^{2} - 3 u) d u

Multiply by the conjugate of

2 u^{2} - 3 u : \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u}

= 2 \cdot \int u^{3} \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u} d u

Simplify

u^{3} \frac{4 u ^{4} - 9 u ^{2}}{2 u ^{2} + 3 u} : u^{4} (2 u - 3)

= 2 \cdot \int u^{4} (2 u - 3) d u

Apply Integration By Parts

: \frac{1}{5} u^{5} (2 u - 3) - \int \frac{2 u ^{5}}{5} d u

= 2 (\frac{1}{5} u^{5} (2 u - 3) - \int \frac{2 u ^{5}}{5} d u)

\int \frac{2 u ^{5}}{5} d u = \frac{u ^{6}}{15}

= 2 (\frac{1}{5} u^{5} (2 u - 3) - \frac{u ^{6}}{15})

Substitute back

u = t

= 2 (\frac{1}{5} (t)^{5} (2 t - 3) - \frac{( t ) ^{6}}{15})

Simplify

2 (\frac{1}{5} (t)^{5} (2 t - 3) - \frac{( t ) ^{6}}{15}) : 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15})

= 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15})

Add a constant to the solution

= 2 (\frac{1}{5} t^{\frac{5}{2}} (2 t - 3) - \frac{t ^{3}}{15}) + C

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

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What is the integral of t(2t-3sqrt(t)) ?
The integral of t(2t-3sqrt(t)) is 2(1/5 t^{5/2}(2sqrt(t)-3)-(t^3)/(15))+C