Integration of Logaritmic Functions

**frog09** · January 2nd 2009, 09:42 AM

$ \int_1^3 (x+3)/(x^2+6x) dx = ? $

I tired separating and simplifying the problem into two different integrals:

$ \int_1^3 1/(x+6)$ + $ \int_1^3 3/(x^2+6x) $

and get...
$\int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x)$

evaluating I get

$ [ln9-ln7]+ 3[ln27+ln7] $

but that is not any of the solutions.

**Scott H** · January 2nd 2009, 09:57 AM

Try Integration by Substitution, which is basically the Chain Rule backward:

$\int\,f'(g(x))g'(x)\,dx\,=\,f(g(x))\,+\,C$

Letting $u\,=\,g(x)\,=\,x^2\,+\,6x$ , we can see that $du\,=\,g'(x)\,dx\,=\,(2x\,+\,6)\,dx\,=\,2(x\,+\,3) \,dx$ , so that the integral becomes

$\int_{u\,=\,7}^{u\,=\,27}\,\frac{1}{2u}\,du.$

Hope this helps.

**frog09** · January 2nd 2009, 10:32 AM

Natural log is supposed to be in the answer...any ideas?

**Isomorphism** · January 2nd 2009, 10:39 AM

Hi frog09,

Do you understand what Scott H says in #2? He has almost completely worked the problem for you.

Cant you figure out what $\int_{u\,=\,7}^{u\,=\,27}\,\frac{1}{2u}\,du$ is ?

If you integrate the above you will see the "supposed natural logs"

**frog09** · January 2nd 2009, 10:57 AM

yes! a moment of enlightenment. thanks!

**DeMath** · January 2nd 2009, 11:05 AM

Originally Posted by frog09

$ \int_1^3 (x+3)/(x^2+6x) dx = ? $

I tired separating and simplifying the problem into two different integrals:

$ \int_1^3 1/(x+6)$ + $ \int_1^3 3/(x^2+6x) $

and get...
$\int_1^3 ln(x+6) + \int_1^3 3ln(x^2+6x)$

evaluating I get

$ [ln9-ln7]+ 3[ln27+ln7] $

but that is not any of the solutions.

$\begin{gathered}\int\limits_1^3 {\frac{{x + 3}}{{x^2 + 6x}}dx} = \frac{1}{2}\int\limits_1^3 {\frac{{2x + 6}}{{x^2 + 6x}}dx = } \frac{1}{2}\int\limits_1^3 {\frac{{d\left( {x^2 + 6x} \right)}}{{x^2 + 6x}} = } \hfill \\= \left. {\frac{1}{2}\ln \left( {x^2 + 6x} \right)} \right|_1^3 = \frac{1}{2}\left( {\ln 27 - \ln 7} \right) = \frac{1}{2}\left( {3\ln 3 - \ln 7} \right). \hfill \\\hfill \\ \end{gathered}$

**Scott H** · January 2nd 2009, 11:20 AM

Are you sure? I thought we had to change the bounds of integration after the change of variables.

Frog09, remember also that $\ln\,u$ is defined as $\int_1^u\,\frac{1}{x}\,dx.$

**tom@ballooncalculus** · January 2nd 2009, 11:49 AM

Except that he didn't actually change the variable, so it was ok. This is how I would keep track of it...

(As usual straight continuous lines diff or anti-diff with respect to x, straight dashed lines with respect to the dashed balloon expression.)

Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers

**DeMath** · January 2nd 2009, 11:52 AM

Originally Posted by Scott H

Are you sure? I thought we had to change the bounds of integration after the change of variables.

Frog09, remember also that $\ln\,u$ is defined as $\int_1^u\,\frac{1}{x}\,dx.$

Yes, I am sure!
In my solution we must not change the bounds of integration.
Because

$\left. {\int\limits_b^a {\frac{{f'\left( x \right)}}{{f\left( x \right)}}dx = } \ln \left( {f\left( x \right)} \right)} \right|_b^a = \ln \left| {f\left( a \right)} \right| - \ln \left| {f\left( b \right)} \right|.$

**Scott H** · January 3rd 2009, 06:36 PM

All right, I see.

Thread: Integration of Logaritmic Functions

LinkBack

Thread Tools

Display

Integration of Logaritmic Functions

Similar Math Help Forum Discussions

logaritmic equation

[SOLVED] Integration of Incomplete Beta Functions and Hypergeometric Functions

[SOLVED] Solve Logaritmic Equation in terms of natural log

Logaritmic equations

integration using trigonmetric functions

Search Tags