Integration problem

[mathwannabe]

Hello everybody

I have been practicing some integrals, but there is one problem that buggs me. Let me show you (steps will be numerated):

(1) \(\displaystyle \int \dfrac{dx}{2x^2+4x+20}=\)

(2) \(\displaystyle \int \dfrac{dx}{2(x^2+2x+10)}=\)

(3) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{x^2+2x+10}=\)

(4) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{(x^2+2x+1)+9}=\)

(5) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{(x+1)^2+9}=\)

(6) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{9(\frac{x+1}{3})^2+1}=\)

(7) \(\displaystyle \frac{1}{18}\int \dfrac{dx}{(\frac{x+1}{3})^2+1}=\) <------ I had no idea how to continue from this step, so I had to look into my notebook for a hint and this is what I found:

(8) \(\displaystyle \frac{1}{18}\int \dfrac{3d(\frac{x+1}{3})}{(\frac{x+1}{3})^2+1}=\) <------ How dis she pull that 3 out from dx when it was not d(x+1) but dx ???

I can finish the problem from here on my own, but I have no idea how she pulled that 3 out form dx...

This will end up being some arctg (arctan) so, that's not a problem...

 

[JeffM]

Hello everybody

I have been practicing some integrals, but there is one problem that buggs me. Let me show you (steps will be numerated):

(1) \(\displaystyle \int \dfrac{dx}{2x^2+4x+20}=\)

(2) \(\displaystyle \int \dfrac{dx}{2(x^2+2x+10)}=\)

(3) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{x^2+2x+10}=\)

(4) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{(x^2+2x+1)+9}=\)

(5) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{(x+1)^2+9}=\)

(6) \(\displaystyle \frac{1}{2}\int \dfrac{dx}{9(\frac{x+1}{3})^2+1}=\)

(7) \(\displaystyle \frac{1}{18}\int \dfrac{dx}{(\frac{x+1}{3})^2+1}=\) <------ I had no idea how to continue from this step, so I had to look into my notebook for a hint and this is what I found:

(8) \(\displaystyle \frac{1}{18}\int \dfrac{3d(\frac{x+1}{3})}{(\frac{x+1}{3})^2+1}=\) <------ How dis she pull that 3 out from dx when it was not d(x+1) but dx ???

I can finish the problem from here on my own, but I have no idea how she pulled that 3 out form dx...

This will end up being some arctg (arctan) so, that's not a problem...

\(\displaystyle u = \dfrac{x + 1}{3} \implies u = \dfrac{1}{3} * (x + 1) \implies \dfrac{du}{dx} = \dfrac{1}{3} \implies dx = 3du.\)

(7) \(\displaystyle u = \dfrac{x + 1}{3} \implies \frac{1}{18}\int \dfrac{dx}{(\frac{x+1}{3})^2+1} = \frac{1}{18} \int \dfrac{3du}{u^2 + 1}.\)

 

[mathwannabe]

\(\displaystyle u = \dfrac{x + 1}{3} \implies u = \dfrac{1}{3} * (x + 1) \implies \dfrac{du}{dx} = \dfrac{1}{3} \implies dx = 3du.\)

(7) \(\displaystyle u = \dfrac{x + 1}{3} \implies \frac{1}{18}\int \dfrac{dx}{(\frac{x+1}{3})^2+1} = \frac{1}{18} \int \dfrac{3du}{u^2 + 1}.\)

Got it! Thanks Jeff

 

[daon2]

You may be able to stop your manipulations when you reach

\(\displaystyle \displaystyle \int \dfrac{dx}{(x+1)^2+3^2}\)

In every text I have encountered the following formula is given (and proved)

\(\displaystyle \displaystyle \int \dfrac{du}{u^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{u}{a})+C\)