##### Calculus questions. See the attachment. Thank you very much

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Sep 29th, 2015

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

hope you understood..please message if you have any doubts...thank you
Sep 29th, 2015

Sep 29th, 2015

i checked by using antiderivative

that is diffrentiating the answer and getting the question

Sep 29th, 2015

can you show step by step please

Sep 29th, 2015

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:

1/31/x4dx1/3x3 (ln(4x))=1/3x^−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)/3x31/9x3

Simplify:

ln(4x)/ x^4dx=−(ln(64x3)+1) / 9x3

ln(4x)/ x^4dx=−(ln(64x3)+1)/9x3+C

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

Sep 29th, 2015

these are the steps to find the answer using integration

Sep 29th, 2015

Great! Thank you

Sep 29th, 2015

you are welcome

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

Sep 29th, 2015

I posted a new questions few minutes ago

Sep 29th, 2015

iam sorry

now , i am going to work.

Sep 29th, 2015

okay

How do I pay you?

Sep 29th, 2015

for which question?

Sep 29th, 2015

for solving this one and for new one I posted

Sep 29th, 2015

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

Sep 29th, 2015

ok. thank you

Sep 29th, 2015

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Sep 29th, 2015
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Sep 29th, 2015
Oct 22nd, 2017
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