How Do You Correctly Solve the Integral of [(3-ln2x)^3]/2x?

[dangish]

question: integral of [(3-ln2x)^3]/2x

my workings:

I let u = 3-ln2x
then du= -2/x dx
so -1/2du = 1/x dx

this leaves me with -(1/2)*integral of u^3/2 du

I take the bottom 2 out to get -(1/2)*(1/2) * integral of u^3 du

which is -1/4 * (u^4)/4

then I sub u into get

-1/4 *(3-ln2x)/4

Which is -1/16 * (3-ln2x)

On my test I only got 8/10 for this question and When I plug this example into Wolfram it gives some wacky answer.

Can someone tell me if I'm right? Thanks

 

[LCKurtz]

question: integral of [(3-ln2x)^3]/2x

my workings:

I let u = 3-ln2x
then du= -2/x dx

No. du = -(1/x)dx. You need the chain rule there.

so -1/2du = 1/x dx

this leaves me with -(1/2)*integral of u^3/2 du

I take the bottom 2 out to get -(1/2)*(1/2) * integral of u^3 du

which is -1/4 * (u^4)/4

then I sub u into get

-1/4 *(3-ln2x)⁴/4

You left off the 4th power. Fix those two mistakes and it is correct.

 

[dangish]

Oops.. the 4th power I did write on my test just missed it there..

and damn I can't believe I got that derivative wrong haha

Thanks