Integration by variable substitution: how t^2/(1+t) becomes t - 1 + 1/(1+t) ?

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Totibob

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Hi! I need some help understanding one step in this solving of an integral.

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I'm assuming it's some basic math I'm overlooking but i'll post it here together with the rest of the solution in case I'm missing something else.

Very thankful for any help, tips or tricks!
 
Totibob said:
Hi! I need some help understanding one step in this solving of an integral.

. . .\(\displaystyle \displaystyle \int_1^4\, \ln\left(1\, +\, \sqrt{\strut x\,}\right)\, dx\)

. . .\(\displaystyle \displaystyle t\, =\, \sqrt{\strut x\,},\quad x\, =\, t^2, \quad dx\, =\, 2t\, dt\)

. . .\(\displaystyle \displaystyle \int_1^2\, 2t\, \ln(1\, +\, t)\, dt\, =\, \Big[t^2\, \ln(1\, +\, t)\Big]\Bigg\rvert_1^2\, -\, \)\(\displaystyle \displaystyle \color{blue}{\int_1^2\, \dfrac{t^2}{1\, +\, t}\, dt}\)

. . . . .\(\displaystyle \displaystyle =\, 4\, \ln(3)\, -\, \ln(2)\, -\, \)\(\displaystyle \displaystyle \color{blue}{\int_1^2\, \left(t\, -\, 1\, +\, \dfrac{1}{1\, +\, t}\right)\, dt}\)

. . . . .\(\displaystyle \displaystyle =\, 4\, \ln(3)\, -\, \ln(2)\, -\, \Big[\dfrac{1}{2}\, (t\, -\, 1)^2\, +\, \ln\big\lvert 1\, +\, t \big\rvert \Big]\Bigg\rvert_1^2\)

. . . . .\(\displaystyle \displaystyle =\, 4\, \ln(3)\, -\, ln(2)\, -\, \left(\dfrac{1}{2}\, +\, \ln(3)\right)\, +\, (0\, +\, \ln(2))\, \)

. . . . .\(\displaystyle \displaystyle =\, 3\, \ln(3)\, -\, \dfrac{1}{2}\)

I'm assuming it's some basic math I'm overlooking but i'll post it here together with the rest of the solution in case I'm missing something else.
Click to expand...
I think the author was thinking of this:

. . .\(\displaystyle \displaystyle t^2\, =\, t^2\, -\, 1\, +\, 1\)

. . . . .\(\displaystyle \displaystyle =\, (t^2\, -\, 1)\, +\, 1\)

. . . . .\(\displaystyle \displaystyle =\, (t\, +\, 1)(t\, -\,1)\, +\, 1\)

The first adding and subtracting doesn't change anything, because we've essentially added zero. But it does allow the factorization. Once we have this, we can restate the original fraction as:

. . .\(\displaystyle \displaystyle \dfrac{t^2}{1\, +\, t}\, =\, \dfrac{(t\, +\, 1)(t\, -\, 1)\, +\, 1}{1\, +\, t}\)

. . . . .\(\displaystyle \displaystyle =\, \dfrac{(t\, +\, 1)(t\, -\, 1)}{1\, +\, t}\, +\, \dfrac{1}{1\, +\, t}\)

. . . . .\(\displaystyle \displaystyle =\, (t\, -\, 1)\, +\, \dfrac{1}{1\, +\, t}\)

How did the author come up with this? He probably learned it from somebody else, just like you now have. ;)
 
stapel said:
I think the author was thinking of this:

. . .\(\displaystyle \displaystyle t^2\, =\, t^2\, -\, 1\, +\, 1\)

. . . . .\(\displaystyle \displaystyle =\, (t^2\, -\, 1)\, +\, 1\)

. . . . .\(\displaystyle \displaystyle =\, (t\, +\, 1)(t\, -\,1)\, +\, 1\)

The first adding and subtracting doesn't change anything, because we've essentially added zero. But it does allow the factorization. Once we have this, we can restate the original fraction as:

. . .\(\displaystyle \displaystyle \dfrac{t^2}{1\, +\, t}\, =\, \dfrac{(t\, +\, 1)(t\, -\, 1)\, +\, 1}{1\, +\, t}\)

. . . . .\(\displaystyle \displaystyle =\, \dfrac{(t\, +\, 1)(t\, -\, 1)}{1\, +\, t}\, +\, \dfrac{1}{1\, +\, t}\)

. . . . .\(\displaystyle \displaystyle =\, (t\, -\, 1)\, +\, \dfrac{1}{1\, +\, t}\)

How did the author come up with this? He probably learned it from somebody else, just like you now have. ;)
Click to expand...

That's probably exactly what they were thinking of. I recognize the method but i haven't seen it for a while. I definitely see how it's very useful for other problems as well so it feels good to learn it again, thanks a bunch Stapel!
 
:idea: Another way to see how one expression comes from another is to work backwards.

If you were to combine t - 1 + 1/(t + 1) into a single ratio, you could then reverse the steps.
 
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