What is the integral of (u^3-ue^{6-3u^2}) ?

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

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integral of (u^3-ue^{6-3u^2})

\int (u^{3} - u e^{6 - 3 u^{2}}) d u

\frac{u ^{4}}{4} + \frac{1}{6} e^{6 - 3 u^{2}} + C

Solution steps

\int u^{3} - u e^{6 - 3 u^{2}} 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 u^{3} d u - \int u e^{6 - 3 u^{2}} d u

\int u^{3} d u = \frac{u ^{4}}{4}

\int u e^{6 - 3 u^{2}} d u = - \frac{1}{6} e^{6 - 3 u^{2}}

= \frac{u ^{4}}{4} - (- \frac{1}{6} e^{6 - 3 u^{2}})

Simplify

= \frac{u ^{4}}{4} + \frac{1}{6} e^{6 - 3 u^{2}}

Add a constant to the solution

= \frac{u ^{4}}{4} + \frac{1}{6} e^{6 - 3 u^{2}} + C

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