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

2 Answers

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))

May 9, 2018

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.!