How do you find the antiderivative of #int (arctan(sqrtx)) dx#?
1 Answer
Explanation:
#I=intarctan(sqrtx)dx#
Integration by parts takes the form
Differentiating
#I=uv-intvdu#
#color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dx#
#color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dx#
Letting
#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
#I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)#
#color(white)I=(x+1)arctan(sqrtx)-sqrtx+C#