How do you find the antiderivative of int (arctan(sqrtx)) dx∫(arctan(√x))dx?
1 Answer
Explanation:
I=intarctan(sqrtx)dxI=∫arctan(√x)dx
Integration by parts takes the form
Differentiating
I=uv-intvduI=uv−∫vdu
color(white)(I)=xarctan(sqrtx)-intx/(2sqrtx(1+x))dxI=xarctan(√x)−∫x2√x(1+x)dx
color(white)(I)=xarctan(sqrtx)-1/2intsqrtx/(1+x)dxI=xarctan(√x)−12∫√x1+xdx
Letting
I=xarctan(sqrtx)-1/2intt/(1+t^2)(2tdt)I=xarctan(√x)−12∫t1+t2(2tdt)
color(white)I=xarctan(sqrtx)-intt^2/(1+t^2)dtI=xarctan(√x)−∫t21+t2dt
color(white)I=xarctan(sqrtx)-int(1+t^2-1)/(1+t^2)dtI=xarctan(√x)−∫1+t2−11+t2dt
color(white)I=xarctan(sqrtx)-int(1+t^2)/(1+t^2)dt+int1/(1+t^2)dtI=xarctan(√x)−∫1+t21+t2dt+∫11+t2dt
color(white)I=xarctan(sqrtx)-intdt+arctan(t)I=xarctan(√x)−∫dt+arctan(t)
color(white)I=xarctan(sqrtx)-t+arctan(t)I=xarctan(√x)−t+arctan(t)
Since
I=xarctan(sqrtx)-sqrtx+arctan(sqrtx)I=xarctan(√x)−√x+arctan(√x)
color(white)I=(x+1)arctan(sqrtx)-sqrtx+CI=(x+1)arctan(√x)−√x+C