Question

Evaluate `int(9e^(9t))/(e^(18t)+11e^(9t)+18)dt`?

Answer 1

See below

Explanation:

Intuition make that we try the change `z=e^(9t)` with this we have

`dz=9e^(9t)dt` and integral changes to

`intdz/(z^2+11z+18)=I`

Now factorize `1/(z^2+11z+18)=1/((z+2)(z+9))=A/(z+2)+B/(z+9)`

We found `A=1/7` and `B=-1/7`

And then `I=1/7intdz/(z+2)-1/7intdz/(z+9)=1/7ln(z+2)-1/7ln(z+9)=1/7ln(e^(9t)+2)-1/7ln(e^(9t)+9)+C`

We can re-arrange applying logarithm laws

`I=1/7ln((e^(9t)+2)/(e^(9t)+9))`

Answer 2

`1/7ln|((e^(9t)+2)/(e^(9t)+2))|+C.`

Explanation:

Observe that, `e^(18t)+11e^(9t)+18=(e^(9t)+9)(e^(9t)+2)`.

So, let, `e^(9t)=x. :. e^(9t)*9dt=dx`.

`:. I=int(9e^(9t))/(e^(18t)+11e^(9t)+18)dt`,

`=int1/{(x+9)(x+2)}dx`,

`=1/7int{(x+9)-(x+2)}/{(x+9)(x+2)}dx`,

`=1/7int{(x+9)/((x+9)(x+2))-(x+2)/((x+9)(x+2))}dx`,

`=1/7int{1/(x+2)-1/(x+9)}dx`,

`=1/7{ln|(x+2)-ln|(x+9)|}`,

`=1/7ln|(x+2)/(x+9)|`.

`rArr I=1/7ln|((e^(9t)+2)/(e^(9t)+2))|+C.`

Enjoy Maths.!