What is expand e^{3x}*(u_{2}(t)(t^3+1)) ?

The solution to expand e^{3x}*(u_{2}(t)(t^3+1)) is e^{3x}u_{2}t^4+e^{3x}u_{2}t

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expand e^{3x}*(u_{2}(t)(t^3+1))

expand

e^{3 x} \cdot (u_{2} (t) (t^{3} + 1))

e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

Solution steps

e^{3 x} (u_{2} (t) (t^{3} + 1))

Remove parentheses:

(a) = a

= e^{3 x} u_{2} t (t^{3} + 1)

Apply the distributive law:

a (b + c) = ab + a c

a = e^{3 x} u_{2} t, b = t^{3}, c = 1

= e^{3 x} u_{2} t t^{3} + e^{3 x} u_{2} t \cdot 1

= e^{3 x} u_{2} t^{3} t + 1 \cdot e^{3 x} u_{2} t

Simplify

e^{3 x} u_{2} t^{3} t + 1 \cdot e^{3 x} u_{2} t : e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

= e^{3 x} u_{2} t^{4} + e^{3 x} u_{2} t

(3 x^{2} - 2) e^{x^{3} + 7 x}

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\frac{\partial}{\partial t} ((u (t, x)) (3 x - 2 t))

- \frac{d}{d s} (\frac{a}{s ( s ^{2} - b )})

- \frac{12 ( 6 + 2 )}{17}

What is expand e^{3x}*(u_{2}(t)(t^3+1)) ?
The solution to expand e^{3x}*(u_{2}(t)(t^3+1)) is e^{3x}u_{2}t^4+e^{3x}u_{2}t