Natural Logarithmic Integration

[lamaclass]

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.

 

[BigGlenntheHeavy]

\(\displaystyle 2nd \ one: \ |x^{2}+1| \ = \ x^{2}+1 \ if \ x^{2} +1 \ \ge \ 0, which \ it \ is.\)

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

\(\displaystyle Ergo, \ note: \ \frac{1}{2}ln|x^{2}+1|+C \ = \ \frac{1}{2}ln(x^{2}+1)+C\)

 

[soroban]

Hello, lamaclass!

\(\displaystyle \int \frac{x^2+2x+3}{x^3 + 3x^2 + 9x}\,dx\)

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

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


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

. . Got it?

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

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

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

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

 

[lamaclass]

Thank you both for the help!!