Integration by parts question

[santeria13]

The question is as follows:

find,



and leave your answer in terms of natural logarithms . Alright, so i know the integral of this is 3ln(1+3x) +4ln(1-4x) . and that thus this equals

ln(1+3x)^3 x (1-4x)^4 but am confused on where to go from there?

 

[Deleted member 4993]

The question is as follows:

find,

View attachment 3159

and leave your answer in terms of natural logarithms . Alright, so i know the integral of this is 3ln(1+3x) +4ln(1-4x) . and that thus this equals

ln(1+3x)^3 x (1-4x)^4 but am confused on where to go from there?

I(x) = ln[(1+3x) (1-4x)]........................edited

Evaluate I(0) and I(0.2)


View attachment 3159


= I(0.2) - I(0)

 

[pka]

The question is as follows: find,

View attachment 3159

and leave your answer in terms of natural logarithms . Alright, so i know the integral of this is 3ln(1+3x) +4ln(1-4x) .

That is incorrect: the anti-derivative is \(\displaystyle \ln(1+3x)-\ln(1-4x)\).

 

[santeria13]

That is incorrect: the anti-derivative is \(\displaystyle \ln(1+3x)-\ln(1-4x)\).

right, so that means when i integrate that function that i divide the 3 and 4 by the derivative of their brackets?

 

[pka]

right, so that means when i integrate that function that i divide the 3 and 4 by the derivative of their brackets?

No that is not correct. \(\displaystyle \displaystyle\int_0^{0.2} {\left( {\frac{3}{{1 + 3x}} + \frac{4}{{1 - 4x}}} \right)dx} = \left. {\ln \left( {\frac{{1 + 3x}}{{1 - 4x}}} \right)} \right|_{x = 0}^{0.2}\)

 

[HallsofIvy]

You understand that this problem has nothing to do with "integration by parts", don't you?

 

[DrPhil]

The question is as follows:

find,

View attachment 3159

and leave your answer in terms of natural logarithms . Alright, so i know the integral of this is 3ln(1+3x) +4ln(1-4x) . and that thus this equals

ln(1+3x)^3 x (1-4x)^4 but am confused on where to go from there?

Other tutors have told you how to do this integral, but it looks to me like you haven't gone back to the beginning to see what the "3" and the "4" do in the solution. [Also, whenever you write an integral you need to identify the variable of integration by including "dx" (or other ). You have a sum of two separate terms:

\(\displaystyle \displaystyle \int_0^{0.2} \left( \frac{3}{1+3x}\right)\ dx + \int_0^{0.2} \left( \frac{4}{1-4x}\right)\ dx\)

Looking at the first term, let \(\displaystyle u = 1+3x,\,\,\,du = 3\ dx\)

Then the "3" is part of du, and the 1st integral is (expressed as indefinite integral, because we haven't calculated the limits in terms of u)

\(\displaystyle \displaystyle \int \frac{du}{u} = \ln(u) = \ln(1 + 3x) \).

Similarly, the "4" in the 2nd term is part of the differential.