Thank you for the opportunity to help you with your question!

answer is

option D. - (1/3x^3 ) ln (4x) - 1/9x^3 +c

Can you, please, explain how you got his answer? Thank you

i checked by using antiderivative

that is diffrentiating the answer and getting the question

can you show step by step please

For the integral ∫(1/x4) ln(4x)dx use integration by parts ∫udv=uv−∫vdu.

Let u=ln(4x) and dv=dx/x4.

Then du=(ln(4x))′dx=dx/x and v=∫1/x4dx=−1/3x3

Integral becomes

∫(1/x4)ln(4x)dx=(ln(4x)⋅(−1/3x3)−∫(−1/3x3)⋅(1/x)dx)=(−∫(−1/3x4)dx−(1/3x3) ln(4x))

Apply constant multiple rule ∫cf(x)dx=c∫f(x)dx with c=−13 and f(x)=1/x4:

−∫(−1/3x4)dx−(1/3x3) ln(4x)=−(−1/3∫(1/x4)dx)−(1/ 3x3)ln(4x)

Apply power rule ∫x^ndx=x^(n+1)/ n+1 (n≠−1) with n=−4:

1/3∫1/x4dx−1/3x3 (ln(4x))=1/3∫x^−4dx−(1/3x3)ln(4x)=(1/3)x(−4+1)/(−4+1)−(1/3x3)ln4x)=(−x^−3/3)/3−(1/3x3)ln(4x)=(1/3)(−1/3x3)−(1/3x3)ln(4x)

Therefore,

∫∫ln(4x)/ x^4dx=−ln(4x)/3x3−1/9x3

Simplify:

∫ln(4x)/ x^4dx=−(ln(64x3)+1) / 9x3

Add the constant of integration:

∫ln(4x)/ x^4dx=−(ln(64x3)+1)/9x3+C

Answer: ∫ln(4x)/ x^4dx=−(ln(64x3)+1)/9x^3+C.

= - (1/3x^3 ) ln (4x) - 1/9x^3 +c.......................answer

these are the steps to find the answer using integration

Great! Thank you

you are welcome

if you have any new questions...please assign it to me

I posted a new questions few minutes ago

iam sorry

now , i am going to work.

may be later i can help you

okay

How do I pay you?

for which question?

for solving this one and for new one I posted

if i get time after work for your new question,i will bid on your question .

only then you pay me because sometimes i wont get time.so it is better you prefer some other tutor for help

thank you

ok. thank you

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