What is expand (2e^x+3x+1)*(5x-ln(x)) ?

The solution to expand (2e^x+3x+1)*(5x-ln(x)) is 10e^xx-2e^xln(x)+15x^2-3xln(x)+5x-ln(x)

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expand (2e^x+3x+1)*(5x-ln(x))

expand

(2 e^{x} + 3 x + 1) \cdot (5 x - ln (x))

10 e^{x} x - 2 e^{x} ln (x) + 15 x^{2} - 3 x ln (x) + 5 x - ln (x)

Solution steps

(2 e^{x} + 3 x + 1) (5 x - ln (x))

Distribute parentheses

= 2 e^{x} \cdot 5 x + 2 e^{x} (- ln (x)) + 3 x \cdot 5 x + 3 x (- ln (x)) + 1 \cdot 5 x + 1 \cdot (- ln (x))

Apply minus-plus rules

+ (- a) = - a

= 2 \cdot 5 e^{x} x - 2 e^{x} ln (x) + 3 \cdot 5 xx - 3 x ln (x) + 1 \cdot 5 x - 1 \cdot ln (x)

Simplify

2 \cdot 5 e^{x} x - 2 e^{x} ln (x) + 3 \cdot 5 xx - 3 x ln (x) + 1 \cdot 5 x - 1 \cdot ln (x) : 10 e^{x} x - 2 e^{x} ln (x) + 15 x^{2} - 3 x ln (x) + 5 x - ln (x)

= 10 e^{x} x - 2 e^{x} ln (x) + 15 x^{2} - 3 x ln (x) + 5 x - ln (x)

\frac{( 3 x - 5 ) ( x + 2 )}{( x - 6 ) ( x - 1 )}

(- 5 x - 4 y = 20) (3)

2 ((2 x - 3) (x + 9))

\frac{d}{d t} ((s) (s - 2))

\frac{300 0 ^{8}}{192}

What is expand (2e^x+3x+1)*(5x-ln(x)) ?
The solution to expand (2e^x+3x+1)*(5x-ln(x)) is 10e^xx-2e^xln(x)+15x^2-3xln(x)+5x-ln(x)