How do you integrate?

Question

#int_0^3dx/((x+2)(sqrt(x+1))#

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Answer 1 · 2018-03-26T22:26:26

The integral equals #2arctan(2) - pi/2~~ 0.644#

We let #u = sqrt(x + 1)#. Then #du = 1/(2sqrt(x + 1))dx -> dx = 2sqrt(x + 1)du#.

#I = int_1^2 2/(x +2) du#

Since #u = sqrt(x +1)#, we can see that #u^2 = x +1# and that #x + 2 = u^2 + 1#.

Therefore,

#I = int_1^2 2/(u^2 + 1)#

#I = [2arctanu]_1^2#

#I = 2arctan2 - 2arctan(1)#

#I = 2arctan(2) - 2(pi/4)#

#I = 2arctan(2) - pi/2#

Hopefully this helps!