Question

I'm having some issues with integration by u-substitution. Solving in general, but also how can I determine a u-value that isn't just guess and check? Here's the problem. Integrate(3(dx))/(x(ln(7x)))

Answer 1

because these are all " ax + bx + c", u dont really need to worry about choosing a u substitution...
if it becomes ax / (2+bx) +c ... then start worrying about u values
When i do it, if there's times or divide involved in the function im integrating, then i start to worry about substitution, if it's simply addition and or subtraction... i dont bother

Answer 2

so this function id'd do sequentially:
integral of 6x^5 = ??

Answer 3

I know how to integrate this without u-substitution, but this one specifically asked for it to be solved in that manner.

Answer 4

To answer your question, \[\int\limits 6x ^{5}=x ^{6}\]

Answer 5

weird... there's no need for chain, product or quotient rule here...

Answer 6

i guess you could substitute u = 6x into the eqn... if you really have to... ???

Answer 7

I'll get a different example. \[\int\limits \frac{ 3dx }{ x(\ln(7x)) }\]

Answer 8

for your first example, after you would have separated out the integral, your first term will be
\(\int 6x^5dx\)
now if you "have to" use substitution, you would put \(\large u= x^6\)
du =6x^5 dx
\(\implies \int u du\)
and after integrating , substitute back u = x^6

for 2nd, you can just use u=ln(7x)

Answer 9

***sorry, you would get \(\large \int du\)

Answer 10

jack! whats your u ?? and du ?

Answer 11

I believe I should choose u=ln(7x) which would make\[du=\frac{ 1 }{ 7x }(7)=\frac{ 1 }{ x }\]

Answer 12

whats your du then ??
you missed a point, take u as ln (7x) only, just as brenar did

Answer 13

yes, that would lead to \(\large \int 3du\)
easy to integrate, right!
thats why we use substitution method in integration....

Answer 14

**\(\large \int (3/u) du\)

Answer 15

I guess I get lost after determining a u and du. I can substitute in, giving me \[\int\limits \frac{ 3dx }{ xu }\] I know that the above statement is incorrect because I've substituted in u, which means I can't have dx in there, but... I'm not sure where to go from here.

Answer 16

Integration by parts?

Answer 17

you already know, that du = (1/x) dx , right ??
\(\huge \int\limits \frac{ 3dx }{ xu } = \int\limits \dfrac{ 3 }{ u }\times \dfrac{dx}{x}=\int\limits \dfrac{ 3 }{ u }\times du\)
got this ?

Answer 18

@hartnn how do you separate out this integral?...

Answer 19

Apologies for being rather thick headed, but where did you get \[\int\limits \frac{ 3 }{ u } \times du\] I understand how you separated my original integral, but not the simplification it would seem.

Answer 20

u= ln 7x
du/dx = 1/ (7x) d/dx (7x) = 1/x
so, du = (1/x) dx
got this ?

Answer 21

Yes.

Answer 22

jhanny, whats your doubt exactly ?

Answer 23

so, 3dx/ xu = (3/u) (1/x) dx = (3/u) du
this ?

Answer 24

im with you on this, caught up, lol.

Answer 25

(3/u) (1/x) dx = (3/u) du

Where did the dx go?

Answer 26

nowhere
"u= ln 7x
du/dx = 1/ (7x) d/dx (7x) = 1/x
so, du = (1/x) dx"

(3/u) (1/x) dx = (3/u) du

Answer 27

omg that's way too cool.

Answer 28

brenar, i think you have in mind that, du = 1/x
its actually, du = (1/x) dx

Answer 29

\[\int\limits \frac{ 3 }{ xln(7x) }=3\ln \left| \ln(7x) \right|\]