Question #31e26

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Answer 1 · 2016-12-04T02:14:50

#intarctan(sqrt(x))dx=(x+1)arctan(sqrt(x))-sqrt(x)+C#

It is difficult to work with the #sqrt(x)# as the argument of #arctan#, so we begin by making a substitution.

Let #t = sqrt(x) => dt = 1/(2sqrt(x))dx#. Then

#intarctan(sqrt(x))dx = int(2sqrt(x)arctan(sqrt(x)))/(2sqrt(x))dx#

#=int2tarctan(t)dt#

Next, we will apply integration by parts. To apply the formula #intudv = uv-intvdu#, we let

#u = arctan(t)# and #dv = 2tdt#
#=> du = 1/(1+t^2)dt# and #v = t^2#

Applying the formula, this gives

#int2tarctan(t)dt = t^2arctan(t)-intt^2/(1+t^2)dt#

Focusing on the remaining integral, we have

#intt^2/(1+t^2)dt = int(1-1/(1+t^2))dt#

#=intdt - int1/(1+t^2)dt#

#=t - arctan(t) + C#

Putting this all together, we get our final result:

#intarctan(sqrt(x))dx = int2tarctan(t)dt#

#= t^2arctan(t)-intt^2/(1+t^2)dt#

#=t^2arctan(t) - t+arctan(t) + C#

#=(t^2+1)arctan(t) - t + C#

#=(x+1)arctan(sqrt(x))-sqrt(x)+C#