Answer:
as we know, the integral of [tex]x^{n}[/tex] is [tex]\frac{x^{n+1} }{n+1}[/tex]
so, the integral of x.[tex]2^{x}[/tex] will be found as follows:
here, we will use a trick called 'integration by parts'
let x = u and [tex]2^{x}[/tex] = v
∫uv dx = u∫v dx - ∫[(du/dx)* ∫v dx] dx
∫x.[tex]2^{x}[/tex] dx = x∫[tex]2^{x}[/tex] - ∫[(dx/dx) * ∫[tex]2^{x}[/tex] dx ] dx
∫x.[tex]2^{x}[/tex] dx = x*[tex]\frac{2^{x+ 1}}{x+ 1}[/tex] - 1 * [tex]\frac{2^{x + 1} }{x + 1}[/tex]
= [tex]\frac{2^{x} }{x + 1}[/tex]( x - 1 )
