Question

Evaluate this Integral.

Answer 1

\[\int\limits_{?}^{?} \frac{ \ln(6x ^{2}) }{ x }\]

Answer 2

Try a u substitution. Let \[u=\ln (6x^{2}), du=\frac{ 12x }{ 6x^{2} }dx =\frac{ 2 }{ x }dx\]

Answer 3

@mebs rewrite:$$\frac{\log(6x^2)}x=\frac{\log(6)+2\log(x)}{x}$$and substitute \(u=\log(x)\) hence \(du=\frac1x\,dx\) to get $$\int\frac{\log(6)+2\log(x)}x\,dx=\int(\log(6)+2u)\,du=u\log(6)+u^2+C$$

Answer 4

Yea that makes sense thanks @oldrin.bataku

Answer 5

rewriting in terms of \(x\) we get $$\log(x)\log(6)+\log(x)^2+C=\log(x)(\log(6)+\log(x))+C=\log(x)\log(6x)+C$$

Answer 6

@pgpilot326's method also works and is probably better considering I assumed \(x\gt0\)

Answer 7

I will try both.. thank you both of you..

Answer 8

Oldrin, i don't think you're method is appropriate since both terms are divided by x. Look it over again.

Answer 9

@pgpilot326 trust me the method is appropriate and works. if you don't believe me take its derivative and compare... both terms are supposed to be divided by \(x\) otherwise my substitution makes no sense.

Answer 10

In both your cases you both divided by x.

pgpilot cancelled x during his substitution

oldrin just divided by x...

Answer 11

$$\int\frac{\log(6x^2)}x\,dx$$let \(u=\log(6x^2)\) and we get \(du=\dfrac{12x}{6x^2}\,dx=\dfrac2x\,dx\) and we get:$$\int\frac{\log(6x^2)}x\,dx=\frac12\int u\,du=\frac14u^2+C_1=\frac14\log^2(6x^2)+C_1$$My antiderivative I found was \(\log(x)\log(6)+\log^2(x)+C_2\).

Answer 12

@mebs no I cancelled it out from the substitution. Reread my derivation.

Answer 13

@mebs @pgpilot326 both solutions are equivalent up to a constant for \(x\gt0\)

Answer 14

\[\int\limits \frac{ \ln \left( 6x^{2} \right)}{ x }dx=\frac{ 1 }{ 2 }\int\limits u du=\frac{ 1 }{ 2 }u + C\]

Answer 15

$$\begin{align*}\frac14\log^2(6x^2)+C_1&=\frac14(\log(6x^2))^2+C_1\\&=\frac14(\log(6)+2\log(x))^2+C_1\\&=\frac14(\log^2(6)+4\log(x)\log(6)+4\log^2(x))+C_1\\&=\frac14\log^2(6)+\log(x)\log(6)+\log^2(x)+C_1\end{align*}$$Notice this looks almost identical to mine only \(C_2=\frac14\log^2(6)+C_1\) hence the solutions are off by a mere constant (absorbed into the constant of integration)

Answer 16

Antiderivatives that differ by only a constant are equivalent because the constant falls out in differentiation anyways ...

Answer 17

so it is \[\frac{ 1 }{ 2 }\ln(6x^{2}) +C\]

Answer 18

The major difference is that his is also defined for \(x\lt0\) whereas mine is only valid for \(x\gt0\).. and yes @mebs that is a valid antiderivative

Answer 19

\[\ln \left( x ^{2} \right) \neq \left( \ln \left( x \right) \right)^{2}\]
unless ln x = 2

Answer 20

@pgpilot326 are you playing stupid or do you actually not know what you're talking about? Nobody claimed \(\log(x^2)=(\log(x))^2\) ... the closest to that was rewriting \(\log^2(x)=(\log(x))^2\) which is a matter of notation

Answer 21

no sorry

Answer 22

here @pgpilot326 if you don't believe me:
http://www.wolframalpha.com/input/?i=1%2F4+log%5E2%286x%5E2%29+-+%28log%28x%29log%286%29%2Blog%5E2%28x%29%29
scroll down to 'alternate form assuming x is positive' also visually observe that for \(x\gt0\) the difference is constant (the blue line)

Answer 23

Aright im closing this...were not having a argument your both correct..

Answer 24

i believe you!

Answer 25

it's the same idea with \(x\) and \(x+2\) both being antiderivatives of \(1\) since \(\frac{d}{dx}[x]=\frac{d}{dx}[x+2]=1\) -- they differ by a constant which doesn't affect the slope (derivative) only the position of the graph

Answer 26

Yes, I know. I made an error in looking at our responses. Please forgive me.