Question

Solve with an integral?
∫ 8^x(1+8^x)^9 dx
Should I put a u=1+8^x ? And if I do how do I find the anti-derivative of that and so on?

Answer 1

to find the anti derivative , write 8 as an exponential with a logarithmn.

\[8 =e^\left\{ \ln 8 \right\}\]

Answer 2

do you recall that \(\frac{d}{dx}b^x=b^x\ln(b)\) ?

Answer 3

or what @Spacelimbus said, but it might be easier to know that
\[\int b^x=\frac{b^x}{\ln(b)}\]

Answer 4

I was thinking of just making u=1+8^x
so du would be 8^x/ln 8 Then somehow go from there?

Answer 5

You might not need a substitution at all if you apply the formulas that @satellite73 suggested, just multiply out and apply the chain rule to the second term.

Answer 6

you could always just multiply out

Answer 7

what @Spacelimbus said
no need for a substitution, keep it simple

Answer 8

Multiply out?

Answer 9

So like 8^x*1 + (8^2x)^9 ?

Answer 10

where does the ^9 come from?

Answer 11

Snap. The problem is:
∫8^x(1+8^x)^9 dx
I totally forgot to add that part.

Answer 12

hehehe, in that case your substitution is a good approach.

Answer 13

What would my du=? then
My u=1+8^x dx
du=8^x/ln 8
dx=?

Answer 14

ok that is wrong i have it backwards

Answer 15

\[u=1+8^x\]
\[du=8^x\ln(8)dx\]
\[\frac{du}{\ln(8)}=8^xdx\] and now you are in good shape

Answer 16

\[\frac{1}{\ln(8)}\int u^9du\] etc

Answer 17

YES THANK YOU.

Answer 18

So my final answer I got was 8^x/ln(8) * 9(1+8^x)^8
Correct or no?