Integral 2^3x * e^2x dx?

andre-l

Can someone please show me what I'm doing wrong here?

u = 2^3x du = 6^3x ln(2) (simplified)

dv = e^2x v = 1/2e^2x

2^3x * 1/2e^2x - integral { 1/2e^2x 6^3x ln(2)}

again u = 1/2 e^2x v = 2^3x

dv = 6^3x ln(2) du =1/4 e^2x

2^3x * 1/2e^2x - 1/2e^2x * 2^3x - integral{2^3x * 1/4e^2x}

so I'll call integral 1/4I

I = 2^3x * 1/2e^2x - 1/2e^2x * 2^3x - 1/4I

5/4I = 2^3x * 1/2e^2x - 1/2e^2x * 2^3x

I = 4/5(2^3x * 1/2e^2x) - 4/5(1/2e^2x * 2^3x)

the solutions page gives me

integral {(8e^2)^xdx = 1 / (2 + 2ln(2)) * (8e^2)^x + C

should I rewrite the original problem as

integral {(2^3)^x * (e^2)^x dx ???????

oh ofcourse 2^3 = 8

so 8^x * e^2x

sorry to waste your time. LOL

john-f

integration by parts :

u = 2^(3x) = e^(x ln 8) ⇨ du = ln 8 e^(x ln 8) dx = (ln 8) 2^(3x) dx

dv = e^(2x) dx ⇨ v = (1/2) e^(2x)

⇔

∫ 2^(3x) e^(2x) dx =

[ 2^(3x) (1/2) e^(2x) ] - ∫ (1/2) e^(2x) (ln 8) 2^(3x) dx

⇔

∫ 2^(3x) e^(2x) dx + (1/2) (ln 8) ∫ e^(2x) 2^(3x) dx =

[ 2^(3x) (1/2) e^(2x) ]

⇔

[ 1 + (ln 8 / 2) ] [ ∫ 2^(3x) e^(2x) dx ] = [ 2^(3x) (1/2) e^(2x) ]

∫ 2^(3x) e^(2x) dx = [ 2^(3x) (1/2) e^(2x) ] / [ 1 + (ln 8 / 2) ]

∫ 2^(3x) e^(2x) dx = [ 2^(3x - 1) e^(2x) ] / [ 1 + (3/2) ln 2 ]

∫ 2^(3x) e^(2x) dx = [ 2^(3x - 1) e^(2x) ] / [ (1/2) (2 + 3 ln 2) ]

∫ 2^(3x) e^(2x) dx = [ 2^(3x) e^(2x) ] / [ 2 + 3 ln 2 ] ⇦ first form of the answer

You can also write :

∫ 8^x e^(2x) dx = [ 8^x e^(2x) ] / [ 2 + 3 ln 2 ]

You choose : 2^(3x) = [ 2^3 ]^x = 8^x and ln 8 = ln 2^3 = 3 ln 2

So you can also write :

∫ 8^x e^(2x) dx = [ 8^x e^(2x) ] / [ 2 + ln 8 ] ⇦ second form of the answer

Verification here :

http://integrals.wolfram.com/index.jsp?expr=2^%283...

*******

Be careful !

you did a mistake when you wrote :

integral {(8e^2)^xdx = 1 / (2 + 2ln(2)) * (8e^2)^x + C

It's not 2 ln 2 but 3 ln 2 because 2 ln 2 = ln 4 and you have ln 8.

Rewritting 8^x e^(2x) as [ 8 ]^x [ e^2 ]^x = [ 8 e^2 ]^x is not an obligation but only an option.

Take a look at the answer of wolfram.com