Integration by Substiution

[Rowallan]

If given;
Integ x/(6+4x) dx

using u = 6 + 4x du =4dx and x =(u - 6)/4

=> Integ x/(6+4x) dx = 1/4.Integ (u - 6).1/4. 1/u du
=1/16.Integ 1-6/u du
=1/16(u-6lnu) +C
=1/16((6+4x)-6ln(6+4x)) +C

Am I nearly there or have I got the wrong end of the stick?

 

[soroban]

Hello, Rowallan!

Your work is correct . . . good for you!

\(\displaystyle \displaystyle \int \frac{x}{6+4x}\,dx\)

Since the numerator and denominator are both linear,
. . nothing is gained by a substitution.
I would solve it like this . . .

. . \(\displaystyle \frac{x}{4x+6} \;=\;\frac{x}{4\left(x + \frac{3}{2}\right)} \;=\;\frac{1}{4}\cdot\frac{(x + \frac{3}{2}) - \frac{3}{2}}{x + \frac{3}{2}} \;=\;\frac{1}{4}\left(1 - \frac{\frac{3}{2}}{x + \frac{3}{2}}\right)\)

\(\displaystyle \text{Then: }\;\int\frac{x\.dx}{4x+6} \;=\;\frac{1}{4}\int\left(1 - \frac{\frac{3}{2}}{x + \frac{3}{2}}\right)dx \;=\; \frac{1}{4}\left(x - \tfrac{3}{2}\ln\left|x + \tfrac{3}{2}\right|\right) + C\)

 

[Rowallan]

Thanks, I follow what you have done, but For give me for being slow but how does

1/16((6+4x)-6ln(6+4x)) +C = 3/8+1/4.x - 3/8.lnI6+4xI +C

equal your answer?

I put it into Mathcad, the computer programme we have to use and it gave;

1/4.x - 3/8.lnI12+8xI +C
I'm confused :?

 

[galactus]

Soroban's solution and MathCad's are the same. Look a little closer. Distribute the 1/4.

Divide by 8, then doesn't 12+8x reduce to x+3/2?.

 

[Rowallan]

1/4.x - 3/8.lnI12+8xI +C

OK, distributing 1/4 gives;

1/4(x - 3/2.lnI12 + 8xI) + C

Does the whole equation not need to be divided by 8, not just the ln part?

:?
I don't understand how using substitution gives such a different answer.
This part of a dynamics question; (I replaced v with x before)

v.dv/dx = -9.81/20(6 + 4v)
Integ v/(6+4v)dv = Integ 0.4905 dx
Integ v/(6+4v)dv = 0.4905x + C

I need to find x in terms of v.