What is expand (2+x)(2x+1/x)^5 ?

The solution to expand (2+x)(2x+1/x)^5 is (2(2x^2+1)^5)/(x^5)+((2x^2+1)^5)/(x^4)

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expand (2+x)(2x+1/x)^5

expand

(2 + x) (2 x + \frac{1}{x})^{5}

\frac{2 ( 2 x ^{2} + 1 ) ^{5}}{x ^{5}} + \frac{( 2 x ^{2} + 1 ) ^{5}}{x ^{4}}

Solution steps

(2 + x) (2 x + \frac{1}{x})^{5}

= (2 x + \frac{1}{x})^{5} (2 + x)

Apply the distributive law:

a (b + c) = ab + a c

(2 x + \frac{1}{x})^{5} (2 + x) = (2 x + \frac{1}{x})^{5} \cdot 2 + (2 x + \frac{1}{x})^{5} x

= (2 x + \frac{1}{x})^{5} \cdot 2 + (2 x + \frac{1}{x})^{5} x

Simplify

(2 x + \frac{1}{x})^{5} \cdot 2 + (2 x + \frac{1}{x})^{5} x : \frac{2 ( 2 x ^{2} + 1 ) ^{5}}{x ^{5}} + \frac{( 2 x ^{2} + 1 ) ^{5}}{x ^{4}}

= \frac{2 ( 2 x ^{2} + 1 ) ^{5}}{x ^{5}} + \frac{( 2 x ^{2} + 1 ) ^{5}}{x ^{4}}

\frac{( s ^{2} + 3 )}{( ( s ^{2} + 2 \cdot s + 5 ) \cdot ( s + 2 ) )}

u (t - \frac{7 π}{2}) \cdot sin (t)

y^{''} (x) - (1 + x) y

(A + B) \cdot B^{^{'}}

(\frac{x}{3} + 4)^{5}

What is expand (2+x)(2x+1/x)^5 ?
The solution to expand (2+x)(2x+1/x)^5 is (2(2x^2+1)^5)/(x^5)+((2x^2+1)^5)/(x^4)