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

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

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integral of 6e^x-x^3(sqrt(x)+1)

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

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

Solution steps

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

Apply u-substitution

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

Expand

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

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

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 u - \int \frac{ln ^{\frac{7}{2}} ( u )}{u} d u - \int \frac{ln ^{3} ( u )}{u} d u

\int 6 d u = 6 u

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

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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