What is expand f(x)In(e^{2x}+5) ?

The solution to expand f(x)In(e^{2x}+5) is ne^{2x}fxI+5nfxI

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expand f(x)In(e^{2x}+5)

expand

f (x) I n (e^{2 x} + 5)

n e^{2 x} f x I + 5 n f x I

Solution steps

f (x) I n (e^{2 x} + 5)

= n f (x) I (e^{2 x} + 5)

Apply the distributive law:

a (b + c) = ab + a c

a = f (x) I n, b = e^{2 x}, c = 5

= f (x) I n e^{2 x} + f (x) I n \cdot 5

= n e^{2 x} f (x) I + 5 n f (x) I

Remove parentheses:

(a) = a

= n e^{2 x} f x I + 5 n f x I

(a^{x + 1} - b^{x + 1}) l e f t (b^{x + 1} + a^{x - 1})

e^{y} (y^{2} - x^{2})

\frac{2 x ^{3} - 3 x ^{2} + 4 x}{( 2 x ^{2} - 3 )}

(a - a^{2} - 1)^{3}

(a - a^{2} - 1)^{2}

What is expand f(x)In(e^{2x}+5) ?
The solution to expand f(x)In(e^{2x}+5) is ne^{2x}fxI+5nfxI