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

Question

Emily Ammerman · Accepted Answer

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

Christopher Agresta · Answer

undefined To find the antiderivative of $\int \arctan(\sqrt{x}) \, dx$, use integration by parts with $u = \arctan(\sqrt{x})$ and $dv = dx$. Differentiate $u$ and integrate $dv$ to find $du$ and $v$. Then apply the integration by parts formula:

$$
\int u \, dv = uv - \int v \, du
$$

Finally, substitute the expressions for $u$, $v$, $du$, and integrate the resulting expression.