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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
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:
∫ln(4x)/ x^4dx=−(ln(64x3)+1) / 9x3
Add the constant of integration:
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
now , i am going to work.
may be later i can help you
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
ok. thank you
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