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

2 Answers

Answer:

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

Answer:

# 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.!