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Integral with ln

Yankel

Active member
Jan 27, 2012
398
Hello

I am trying to solve the integral of the function:

3x/(4x-1)

Need some help with it, I am stuck...

Thanks !
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello

I am trying to solve the integral of the function:

3x/(4x-1)

Need some help with it, I am stuck...

Thanks !
\(\displaystyle \displaystyle \begin{align*} \int{\frac{3x}{4x-1}\,dx} &= \frac{3}{4}\int{\frac{4x}{4x-1}\,dx} \\ &= \frac{3}{4}\int{\frac{4x-1 + 1}{4x - 1}\,dx} \\ &= \frac{3}{4}\int{1 + \frac{1}{4x-1}\,dx} \\ &= \frac{3}{4} \left( x + \frac{1}{4}\ln{ | 4x - 1 | } \right) + C \end{align*}\)
 

paulmdrdo

Active member
May 13, 2013
386
\(\displaystyle \displaystyle \begin{align*} \int{\frac{3x}{4x-1}\,dx} &= \frac{3}{4}\int{\frac{4x}{4x-1}\,dx} \\ &= \frac{3}{4}\int{\frac{4x-1 + 1}{4x - 1}\,dx} \\ &= \frac{3}{4}\int{1 + \frac{1}{4x-1}\,dx} \\ &= \frac{3}{4} \left( x + \frac{1}{4}\ln{ | 4x - 1 | } \right) + C \end{align*}\)
did you use partial fraction here?
 

paulmdrdo

Active member
May 13, 2013
386
letting u = 4x-1, u+1/4 = x
getting the derivative of u, du = 4dx, dx = 1du/4

now plug those values
∫{[3(u+1)/4]/4u]}du
∫[(3u+3)/16u]du
3/16(∫du+∫1/u*du)

3/16*u+3/16*ln|u|+c

3/16*(4x-1)+3/16*ln|4x-1| --- this is what i get when i use substitution method. am I correct?
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
did you use partial fraction here?
Yes, it can be thought of as a partial fraction decomposition of \(\frac{4x}{4x-1}\). An easy method to find the partial fractions in this case is the technique used by Prove It; writing the numerator as \(4x-1+1\).

letting u = 4x-1, u+1/4 = x
getting the derivative of u, du = 4dx, dx = 1du/4

now plug those values
∫{[3(u+1)/4]/4u]}du
∫[(3u+3)/16u]du
3/16(∫du+∫1/u*du)

3/16*u+3/16*ln|u|+c

3/16*(4x-1)+3/16*ln|4x-1| --- this is what i get when i use substitution method. am I correct?
Yes, your method is perfect. Note that after a few algebraic simplifications you arrive at the same result obtained by Prove It. :)

\[\frac{3}{16}(4x-1)+\frac{3}{16}\ln|4x-1|+C=\frac{3}{4}\left(x+\frac{1}{4}\ln|4x-1|\right)+C-\frac{3}{16}\]

\(A=C-\frac{3}{16}\) is an arbitrary constant. Therefore,

\[\frac{3}{16}(4x-1)+\frac{3}{16}\ln|4x-1|+C=\frac{3}{4}\left(x+\frac{1}{4}\ln|4x-1|\right)+A\]
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
What I would have done, equivalent to what Prove It and Sudharaka did, is let u= 4x- 1, the denominator. Then 4x= u+ 1 and x= (u+1)/4 and du= 4dx so that dx= (1/4)du.

The integral becomes
[tex]\int \frac{3\frac{u+1}{4}}{u}(du/4)[/tex][tex]= \frac{3}{16}\int \frac{u+1}{u}du[/tex][tex]= \frac{3}{16}\int 1+ \frac{1}{u}du[/tex]