Graph!! Please help !?

1)
http://www.wolframalpha.com/input/?i=graph+x+(ln+x)%5E2+from+0+to+20
f(x) = x (ln x)^2

No horizontal or vertical asymptotes
f'(x) = (ln x)^2 + x (2 ln x) (1/x)
f'(x) = (ln x)^2 + 2 ln x
f'(x) = 0
(ln x)( ln x + 2) = 0
ln x = 0
x = 1
ln x + 2 = 0
ln x = -2
x = e^(-2)
x= 1/e^2 are the critical points
f''(x) = 2 ln x (1/x) + 2/x
when x=1/e^2, f''(x) = -2e^2 < 0 so f has a local maximum at x=1/e^2
Local maximum s f(1/e^2) = (1/e^2) (ln (1/e^2))^2= 4/e^2

minimum occurs when x=1 and the global minimum is 0

Point of inflection:
f''(x) = 2 ln x (1/x) + 2/x
f''(x) = 2 ln x / x + 2/ x = 0
(1/x) (2 ln x + 2) =0
2 ln x = -2
ln x = -1
x = e^(-1)
x = 1/e is the point of inflection

Consider the intervals (0 , 1/e ) and (1/e, ∞)

choose any one point from each interval and evaluate f''(x)
(0 , 1/e ) : choose x=0.1
f''(x) = 2 ln x / x + 2/ x
f''(0.1) = -26.05 < 0 so f is concave down on (0, 1/e)

(1/e, ∞): choose x=1
f''(x) = 2 ln x / x + 2/ x
f''(1) = 2 > 0 so f is concave up on (1/e, ∞)

2)
f(x) = arctan(1+1/x)

Lim x-->infinity f(x) = arctan(1) = pi/4
y=pi/4 is the horizontal asymptote

No vertical asymptote

f(x) = arctan( (x+1)/x)
Let u=(x+1)/x
du/dx = (x(1)-(x+1)(1)) /x^2 = -1/x^2
f'(x) = (1/(1+u^2) ) du/dx
f'(x) = 1 / (1 + (x+1)^2 /x^2) (-1/x^2)
f'(x) = x^2 / ( x^2+ x^2+2x+1) (-1/x^2)
f'(x) = - 1/ (2x^2+2x+1)
f'(x)=0

no solution, no critical point
no minimum or maximum

Point of inflection:
f'(x) = -(2x^2+2x+1)^(-1)
f''(x) = (-1)(-1) (2x^2+2x+1)^(-2) ( 4x+2)
f''(x) = (4x+2) /(2x^2+2x+1)^2 = 0
4x+2 = 0
4x=-2
x=-1/2 (point of inflection)

Consider (-infinity, -1/2) , (-1/2, infinity)
choose any one point from each interval and examine f''(x)
(-infinity, -1/2) :
choose x=-1
f''(-1) = -2 < 0 so f is concave down on (-infinity, -1/2)

(-1/2, infinity)
choose x=1
f''(1) = 6/5 > 0 so f is concave up on (-1/2, infinity)