How do you find the antiderivative of #int (arctan(sqrtx)) dx#?

1 Answer
mason m
Nov 7, 2016

#(x+1)arctan(sqrtx)-sqrtx+C#

Explanation:

#I=intarctan(sqrtx)dx#

Integration by parts takes the form #intudv=uv-intvdu#. So, for #intarctan(sqrtx)dx#, we should let #u=arctan(sqrtx)# and #dv=dx#.

Differentiating #u# gives #(du)/dx=1/(1+(sqrtx)^2)*d/dxsqrtx# or #du=1/(2sqrtx(1+x))dx#. From #dv=dx# we integrate to show that #v=x#. Thus:

#I=uv-intvdu#

#color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dx#

#color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dx#

Letting #t=sqrtx# implies that #t^2=x# and furthermore, that #2tdt=dx# through differentiating. Thus:

#I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)#

#color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dt#

#color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dt#

#color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dt#

#color(white)I=xarctan(sqrtx)-intdt+arctan(t)#

#color(white)I=xarctan(sqrtx)-t+arctan(t)#

Since #t=sqrtx#:

#I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)#

#color(white)I=(x+1)arctan(sqrtx)-sqrtx+C#

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