What is the integral of [6e^u-u^3(sqrt(u)+1)] ?

The integral of [6e^u-u^3(sqrt(u)+1)] is 6e^u-2/9 u^{9/2}-(u^4)/4+C

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integral of [6e^u-u^3(sqrt(u)+1)]

\int [6 e^{u} - u^{3} (u + 1)] d u

6 e^{u} - \frac{2}{9} u^{\frac{9}{2}} - \frac{u ^{4}}{4} + C

Solution steps

\int 6 e^{u} - u^{3} (u + 1) d u

Apply u-substitution

: \int \frac{6 v - ln ^{3} ( v ) ( ln ( v ) + 1 )}{v} d v

= \int \frac{6 v - ln ^{3} ( v ) ( ln ( v ) + 1 )}{v} d v

Expand

\frac{6 v - ln ^{3} ( v ) ( ln ( v ) + 1 )}{v} : 6 - \frac{ln ^{\frac{7}{2}} ( v )}{v} - \frac{ln ^{3} ( v )}{v}

= \int 6 - \frac{ln ^{\frac{7}{2}} ( v )}{v} - \frac{ln ^{3} ( v )}{v} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 6 d v - \int \frac{ln ^{\frac{7}{2}} ( v )}{v} d v - \int \frac{ln ^{3} ( v )}{v} d v

\int 6 d v = 6 v

\int \frac{ln ^{\frac{7}{2}} ( v )}{v} d v = \frac{2}{9} ln^{\frac{9}{2}} (v)

\int \frac{ln ^{3} ( v )}{v} d v = \frac{ln ^{4} ( v )}{4}

= 6 v - \frac{2}{9} ln^{\frac{9}{2}} (v) - \frac{ln ^{4} ( v )}{4}

Substitute back

v = e^{u}

= 6 e^{u} - \frac{2}{9} ln^{\frac{9}{2}} (e^{u}) - \frac{ln ^{4} ( e ^{u} )}{4}

Simplify

6 e^{u} - \frac{2}{9} ln^{\frac{9}{2}} (e^{u}) - \frac{ln ^{4} ( e ^{u} )}{4} : 6 e^{u} - \frac{2}{9} u^{\frac{9}{2}} - \frac{u ^{4}}{4}

= 6 e^{u} - \frac{2}{9} u^{\frac{9}{2}} - \frac{u ^{4}}{4}

Add a constant to the solution

= 6 e^{u} - \frac{2}{9} u^{\frac{9}{2}} - \frac{u ^{4}}{4} + C

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What is the integral of [6e^u-u^3(sqrt(u)+1)] ?
The integral of [6e^u-u^3(sqrt(u)+1)] is 6e^u-2/9 u^{9/2}-(u^4)/4+C