How do you evaluate #\int \tan ^ { - 1} \sqrt { x } d x#?

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Answer 1 · 2017-12-07T14:17:51

#int arctan(sqrtx)*dx=(x+1)*arctan(sqrtx)-sqrtx+C#

#int arctan(sqrtx)*dx#

After using #u=sqrtx#, #x=u^2# and #dx=2udu# transforms, tis integral became

#int 2u*arctanu*du#

=#u^2*arctanu-int u^2*(du)/(u^2+1)#

=#u^2*arctanu-int (u^2*du)/(u^2+1)#

=#u^2*arctanu-int ((u^2+1-1)*du)/(u^2+1)#

=#u^2*arctanu-int du+int (du)/(u^2+1)#

=#u^2*arctanu-u+arctanu+C#

=#(u^2+1)*arctanu-u+C#

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