Math help please ( calc 2)?

Question

I'm not even sure of how to go about this problem, I tried u sub and integration by parts but I'm not getting the right answer, I don't want the right answer I just want someone to tell me the method

integral of (8e^(x)+2)/(e^(x)) dx

Answer 1

You can use u substitution and then division, or just skip straight to the division:



∫ (8eˣ+2) / (eˣ) dx

∫ (8eˣ) / (eˣ) + 2 / (eˣ) dx

∫ 8 + 2e⁻ˣ dx

8x - 2e⁻ˣ + C

Answer 2

integral (2+8 e^x)/e^x dx



For the integrand (2+8 e^x)/e^x, substitute u = e^x and du = e^x dx:

= integral (2+8u)/u^2 du



For the integrand (2+8u)/u^2, substitute s = 2+8u and ds = 8 du:

= 8 integral s/(-2+s)^2 ds



Expanding the integrand s/(-2+s)^2 gives s/(4-4 s+s^2):

= 8 integral s/(4-4 s+s^2) ds



Rewrite the integrand s/(4-4 s+s^2) as 2/(4-4 s+s^2)+(-4+2 s)/(2 (4-4 s+s^2)):

= 8 integral (2/(4-4 s+s^2)+(-4+2 s)/(2 (4-4 s+s^2))) ds



Integrate the sum term by term and factor out constants:

= 16 integral 1/(4-4 s+s^2) ds+4 integral (-4+2 s)/(4-4 s+s^2) ds



For the integrand (-4+2 s)/(4-4 s+s^2), substitute p = 4-4 s+s^2 and dp = (-4+2 s) ds:

= 4 integral 1/p dp+16 integral 1/(4-4 s+s^2) ds



For the integrand 1/(4-4 s+s^2), complete the square:

= 16 integral 1/(-2+s)^2 ds+4 integral 1/p dp



For the integrand 1/(-2+s)^2, substitute w = -2+s and dw = ds:

= 16 integral 1/w^2 dw+4 integral 1/p dp



The integral of 1/w^2 is -1/w:

= -16/w+4 integral 1/p dp



The integral of 1/p is log(p):

= -16/w+4 log(p)+constant



Substitute back for w = -2+s:

= (4 (-4+(-2+s) log(p)))/(-2+s)+constant



Substitute back for p = 4-4 s+s^2:

= (4 (-4+(-2+s) log((-2+s)^2)))/(-2+s)+constant



Substitute back for s = 2+8 u:

= -2/u+4 log(64 u^2)+constant



Substitute back for u = e^x:

= -2 e^(-x)+4 log(64 e^(2 x))+constant, which is equivalent for restricted x values to:

= -2 e^(-x)+8 x+constant and that's your answer



J

Answer 3

You can split the integrand up as follows:



∫ [(8e^x / e^x) + (2/ e^x) ]dx = ∫ [8 + 2e^(-x) ]dx

Answer 4

(8e^(x)+2)/(e^(x)) = 8 + 2e^(-x)



Integrate 8+ 2 e^(-x)

= 8x - 2e^(-x) + C

Answer 5

put e^x =t

Differentiate e^x *dx =dt

dx = dt/e^x =dt/t

integrate[(8t+2)/t^2] dt

integrate[8/t +2/t^2]

8 log (t) -2/t + C .........................Ans