integration?

Question

integrate: arctan(1/x) dx

Answer 1

Make a substitution:
u = 1/x, x = 1/u, dx = -1/u² du
We get
-∫ arctan u du/u²
Now use integration by parts
U = arctan u dV = du/u²
dU = du/ u²+1 V = -1/u
So we have( because of the - sign)
arctan u/u - ∫ du/u(u²+1)
Third technique in the same problem! Now use
partial fractions to do the last integral.
1/u(u²+1) = A/u + (Bu+C)/(u²+1)
1 = A(u²+1) + (Bu+C)u
u= 0 yields A = 1.
Equating coefficients of like terms,
A + B = 0, B = -1
C = 0
So, finally, the last integral yields,
∫ 1/u -u/(u²+1) du
= log u - ½ log(u²+1)
So our final result is
arctan u/u -log u +½ log(u²+1).
Now back-substitute to finish the job:
∫ arctan(1/x) dx = x arctan(1/x) + log x + ½ log( (x²+1)/x² ) + C.
= x arctan(1/x) + ½ log(x²+1) + C.
A tough integral, but worth the effort, I think!

Answer 2

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