Evaluate int(9e^(9t))/(e^(18t)+11e^(9t)+18)dt9e9te18t+11e9t+18dt?

2 Answers

See below

Explanation:

Intuition make that we try the change z=e^(9t)z=e9t with this we have

dz=9e^(9t)dtdz=9e9tdt and integral changes to

intdz/(z^2+11z+18)=Idzz2+11z+18=I

Now factorize 1/(z^2+11z+18)=1/((z+2)(z+9))=A/(z+2)+B/(z+9)1z2+11z+18=1(z+2)(z+9)=Az+2+Bz+9

We found A=1/7A=17 and B=-1/7B=17

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)+CI=17dzz+217dzz+9=17ln(z+2)17ln(z+9)=17ln(e9t+2)17ln(e9t+9)+C

We can re-arrange applying logarithm laws

I=1/7ln((e^(9t)+2)/(e^(9t)+9))I=17ln(e9t+2e9t+9)

May 9, 2018

1/7ln|((e^(9t)+2)/(e^(9t)+2))|+C.17ln(e9t+2e9t+2)+C.

Explanation:

Observe that, e^(18t)+11e^(9t)+18=(e^(9t)+9)(e^(9t)+2)e18t+11e9t+18=(e9t+9)(e9t+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.!