How do you integrate #e^sqrt(4x + 9)#?

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Answer 1 · 2017-09-05T01:04:33

The integral equals # 1/2sqrt(4x + 9)e^sqrt(4x + 9) - 1/2e^sqrt(4x + 9) + C#

A logical first step would be to let #u = 4x + 9#. Then #du = 4dx# and the integral becomes:

#I = 1/4int e^sqrt(u)du#

If we now let #t = sqrt(u)#, then #dt = 1/(2sqrt(u)) du# and #2sqrt(u)dt = du#

#I = 1/4int 2te^tdt#

#I = 1/2int te^t dt#

So now we can use integration by parts. Letting

#{(n = t), (dm = e^tdt):}#

We get

#{(dn = dt), (m = e^t):}#

We have:

#I = 1/2(te^t - int e^t)#

#I = 1/2te^t - 1/2e^t + C#

Reversing the substitutions:

#I = 1/2sqrt(u)e^sqrt(u) - 1/2e^sqrt(u) + C#

#I = 1/2sqrt(4x + 9)e^sqrt(4x + 9) - 1/2e^sqrt(4x + 9) + C#

Hopefully this helps!