Natural Logarithmic Integration

lamaclass

Junior Member
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Oct 18, 2009
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69
I don't understand this problem:

IN [x[sup:3d9qnysh]2[/sup:3d9qnysh]+2x+3/x[sup:3d9qnysh]3[/sup:3d9qnysh]3x[sup:3d9qnysh]2[/sup:3d9qnysh]+9x dx]

Here's what I've figured out:

u=x[sup:3d9qnysh]3[/sup:3d9qnysh]+3x[sup:3d9qnysh]2[/sup:3d9qnysh]+9x

du=3x[sup:3d9qnysh]2[/sup:3d9qnysh]+6x+9

=3(x[sup:3d9qnysh]2[/sup:3d9qnysh]+2x+3 dx)

dx=1/3 du

=1/3 IN [3(x[sup:3d9qnysh]2[/sup:3d9qnysh]+2x+3)/x[sup:3d9qnysh]3[/sup:3d9qnysh]+3x[sup:3d9qnysh]2[/sup:3d9qnysh]+9x) dx]

How would you continue on after this point? :?

Also for this problem:

IN [x/x[sup:3d9qnysh]2[/sup:3d9qnysh]+1 dx]

What I did:

u=x[sup:3d9qnysh]2[/sup:3d9qnysh]+1

du=2xdx

xdx=1/2du

=1/2 IN [1/u du]

=1/2 ln [x[sup:3d9qnysh]2[/sup:3d9qnysh]+1]+c

My book says there would not be an absolute value in the answer and I was not sure why.
 
2nd one: x2+1 = x2+1 if x2+1  0,which it is.\displaystyle 2nd \ one: \ |x^{2}+1| \ = \ x^{2}+1 \ if \ x^{2} +1 \ \ge \ 0, which \ it \ is.

x2+1 = (x2+1) if x2+1 < 0, Impossible.\displaystyle |x^{2}+1| \ = \ -(x^{2}+1) \ if \ x^{2}+1 \ < \ 0, \ Impossible.

Ergo, note: 12lnx2+1+C = 12ln(x2+1)+C\displaystyle Ergo, \ note: \ \frac{1}{2}ln|x^{2}+1|+C \ = \ \frac{1}{2}ln(x^{2}+1)+C
 
Hello, lamaclass!


Let: u=x3+3x2+9xdu=(3x2+6x+9)dx\displaystyle \text{Let: }\,u \,=\,x^3+3x^2 + 9x \qquad\Rightarrow\quad du \,=\,(3x^2 + 6x + 9)dx

. .. . . . du=3(x2+2x+3)dx(x2+2x+3)dx=13du\displaystyle \Rightarrow\quad du \:=\:3(x^2+2x+3)dx \quad\Rightarrow\quad (x^2+2x+3)dx \:=\:\tfrac{1}{3}du


Substitute:   (x2+2x+3)dxThis is 13dux3+3x2+9xThis is u  =  13duu  =  13duu\displaystyle \text{Substitute: }\;\int\frac{\overbrace{(x^2+2x+3)dx}^{\text{This is }\frac{1}{3}du}}{\underbrace{x^3+3x^2+9x}_{\text{This is }u}} \;=\;\int\frac{\frac{1}{3}du}{u} \;=\;\frac{1}{3}\int\frac{du}{u}

. . Got it?




xx2+1dx12lnx2+1+C\displaystyle \int\frac{x}{x^2+1}\,dx \quad\Rightarrow\quad \tfrac{1}{2}\ln|x^2+1| + C . . . . right!

My book says there would not be an absolute value in the answer and I was not sure why.

\(\displaystyle \text{Since }\,x^2\,\geq \,0,\,\text{ then: }\,x^2+1\,\geq\,0 \quad\hdots\quad \text{That is: }\:x^2+1\text{ is never negative}\)

. . So the absolute value "bars" are unnecessary.\displaystyle \text{So the absolute value "bars" are unnecessary.}

 
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