Definite Integration

**Bashyboy** · April 28th 2012, 12:16 PM

The problem is $\int_1^e\frac{(\1+lnx)^2}{x}dx$

So, I figured that $u\equiv\1+lnx$ ,

then $\frac{du}{dx} = \frac{1}{x}\rightarrow{dx} = {x}{du}$

coming upon this step I am not entirely confident with what I am suppose to be substituting.

**Plato** · April 28th 2012, 12:31 PM

Originally Posted by Bashyboy

The problem is $\int_1^e\frac{(\1+lnx)^2}{x}dx$
So, I figured that $u\equiv\1+lnx$ ,
then $\frac{du}{dx} = \frac{1}{x}\rightarrow{dx} = {x}{du}$
coming upon this step I am not entirely confident with what I am suppose to be substituting.

$\mathop {\left. {\frac{{{{\left( {1 + \ln (x)} \right)}^3}}}{3}} \right|}\nolimits_{x = 1}^{x = e}$

**MathoMan** · April 28th 2012, 12:37 PM

Setting $u=1+\ln x$ then differential of the left side of the equation should equal the differential of the right side so $du=d(1+\ln x)=(1+\ln x)'dx=\frac{dx}{x}$ and you should do the following:

$\int\limits_1^c\frac{(1+\ln x)^2}{x}dx=\int\limits_1^c (1+\ln x)^2\frac{dx}{x}=\left\vert\begin{array}{l}1+\ln x=u \\ \frac{dx}{x}=du\\ 1\to 1+\ln 1=1\\ c\to 1+\ln c \end{array}\right\vert=\int\limits_1^{1+\ln c}u^2 du$

**Bashyboy** · April 28th 2012, 12:41 PM

Okay, so I was on the correct path. I ended up with $\int_1^e\frac{u^2}{x}{x}{du}$ ,
but wasn't sure if it was it was correct; now, obviously, it is correct, but I can't seem to figure out why it is correct. I am just trying to justify every step, but it just seems unusual that the x's cancel out and stuff like that.

**Bashyboy** · April 28th 2012, 12:49 PM

Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."

**Plato** · April 28th 2012, 12:57 PM

Originally Posted by Bashyboy

Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."

Did you see reply #2?
Try to get away from using u-substitution. Learn the forms differentiation takes on.

**Bashyboy** · April 28th 2012, 01:03 PM

Plato, are you speaking of your reply, because I am not really sure how you arrived at your answer.

**Plato** · April 28th 2012, 01:06 PM

Originally Posted by Bashyboy

Plato, are you speaking of your reply, because I am not really sure how you arrived at your answer.

What is the derivative of $y=\frac{(1+\ln(x))^3}{3}~?$

Learn the basic derivative forms.

**BrownianMan** · April 28th 2012, 01:18 PM

Originally Posted by Bashyboy

Mathoman, I am not exactly sure what is meant by "then differential of the left side of the equation should equal the differential of the right side."

He simply means that given u = 1 + ln x, the differential of u must be equal to the differential of 1 + ln x. The differential u is du, and the differential of 1 + ln x is dx/x. So when you do the substitution, you end up with the integral of u^2 from 1 to 1+ln e.

**skeeter** · April 28th 2012, 01:21 PM

$\int_1^e \frac{(1+\ln{x})^2}{x} \, dx = \int_1^e \color{red}{(1+\ln{x})^2} \cdot \color{blue}{\frac{1}{x} \, dx} \, \color{black}{= \int_1^2} \color{red}{u^2} \, \color{blue}{du}$

**Prove It** · April 28th 2012, 08:15 PM

Originally Posted by Plato

Did you see reply #2?
Try to get away from using u-substitution. Learn the forms differentiation takes on.

Plato, you should not discourage a student from using a correct method (especially one that is very straightforward, such as a u substitution), especially if it's the method the student has been told to use. Yes, it would be nice if a student can recognise what particular derivatives look like, and therefore what their antiderivatives look like, but that's an unrealistic expectation, as it requires students to remember far too much. It's much better to have students remember a single process than an uncountable number of derivatives.

**Bashyboy** · April 29th 2012, 03:16 AM

I figured by doing the u-substitution I would have one sure-fire way of doing, and that I would also see how the derivative and integrals relate in the process; so, now, I am just trying to make sense of the u-substitution in each integration case.

**Prove It** · April 29th 2012, 04:19 AM

Originally Posted by Bashyboy

I figured by doing the u-substitution I would have one sure-fire way of doing, and that I would also see how the derivative and integrals relate in the process; so, now, I am just trying to make sense of the u-substitution in each integration case.

Well like I said, a good strategy is to look for an inner function and see if its derivative is a multiple.

In this case, you have $\displaystyle \begin{align*} \frac{\left(1 + \ln{x}\right)^2}{x} = \left(1 + \ln{x}\right)^2\frac{1}{x} \end{align*}$

Can you see that if we use $\displaystyle \begin{align*} 1 + \ln{x} \end{align*}$ as the inner function, its derivative is $\displaystyle \begin{align*} \frac{1}{x} \end{align*}$ ? So the integrand is of the form we require.

So we'd let $\displaystyle \begin{align*} u = 1 + \ln{x} \implies \frac{du}{dx} = \frac{1}{x} \implies du = \frac{1}{x}\,dx \end{align*}$ , and note that when $\displaystyle \begin{align*} x = 1, u = 1 \end{align*}$ and when $\displaystyle \begin{align*} x = e, u = 2 \end{align*}$ , then the integral becomes

$\displaystyle \begin{align*} \int_1^e{\left(1 + \ln{x}\right)^2\frac{1}{x}\,dx} &= \int_1^2{u^2\,du} \end{align*}$

which is very easy to solve

**Ivanator27** · April 30th 2012, 10:13 AM

Another way to go about it:

$\int_{1}^{e}\frac{(1+lnx)^{2}}{x}dx$

let $u = 1 + lnx$
$du = \frac{1}{x}dx$
$xdx = du$

Integrating indefinitely yields:

$\int \frac{(u)^{2}}{x}xdu = \int u^2 du = \frac{1}{3}u^3+c$

So...
$\int_{1}^{e}\frac{(1+lnx)^{2}}{x}dx$
$=\left [ \frac{1}{3}(1+lnx)^{3}+c \right ]_{1}^{e}$
And because the Cs cancel anyway:
$=\left [ \frac{1}{3}(1+lnx)^{3}\right ]_{1}^{e}$

This is how I've been taught to approach an example like this, I only change the upper and lower limits if substituting x back into the expression would make it unmanageable messy. Also, it's not so uncommon for all the Xs to cancel out of an example - when I learned how to integrate by substitution all the examples simplified greatly with the correct substitution.

**tom@ballooncalculus** · April 30th 2012, 01:28 PM

And, just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

_________________________________________

Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

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