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Integral from 0 to ln 3 of #e^(3x)/(e^(6x)+5)# . Help please?

Answer 1

Emma Aldridge

See process below

#I=inte^(3x)/(e^(6x)+5)dx#

Change of variable #t=e^(3x)#, then #dt=3e^(3x)dx=3tdx#

With this change we have

#I=intcancelt/(t^2+5)·1/(3cancelt)dt=1/3int1/(t^2+5)#

Other change of variable #t=sqrt5tantheta#. Then #dt=sqrt5sec^2theta# and integral

#I=1/3int1/(5tan^2theta+5)·sqrt5sec^2thetad theta=#

#=1/3·sqrt5/5int(cancelsec^2thetad theta)/(cancelsec^2theta)=#

#=sqrt5/15intd theta=sqrt5/15theta=sqrt5/15arctan(t/sqrt5)=#

#=sqrt5/15arctan(e^(3x)/sqrt5)=F(x)#

#F(ln3)-F(0)=sqrt5/15arctan(27/sqrt5)-sqrt5/15arctan(1/sqrt5)#

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Answer 2

Christopher Allers

To solve the integral ∫(0 to ln 3) e^(3x)/(e^(6x) + 5), first make a substitution: Let u = e^(3x), then du = 3e^(3x)dx.

Now rewrite the integral in terms of u: ∫(0 to ln 3) e^(3x)/(e^(6x) + 5) dx = (1/3) ∫(1 to 3) 1/(u^2 + 5) du.

Now, using the arctangent integral formula, we have: ∫ 1/(u^2 + a^2) du = (1/a) arctan(u/a) + C

Thus, ∫(1 to 3) 1/(u^2 + 5) du = (1/sqrt(5)) * [arctan(3/sqrt(5)) - arctan(1/sqrt(5))].

So, the integral evaluates to: (1/3) * (1/sqrt(5)) * [arctan(3/sqrt(5)) - arctan(1/sqrt(5))].

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.