Evaluate the Integral, ∫tanh^-1(x)dx A.) by using the identity tan^-1 x = 1/2 ln((1+x/)(1-x)). B.) evaluate using parts. C.) Show that both answers are equal up to a constant.

Answer 1

For ease notation I will write

arctanh instead of tanh^-1

A.

After fixing the typos:

∫arctanh(x)dx =

∫(1/2) ln((1+x)/(1-x))dx

By parts using the formula ∫u'v = uv - ∫uv' where

u' = 1/2, v = ln((1+x)/(1-x))

u = x/2,

v' = ((1-x)/(1+x)) (1(1-x) - (-1)(1+x))/(1-x)²

= ((1-x)/(1+x)) 2/(1-x)²

= 2/((1+x)(1-x))

= 2/(1-x²)

Then

∫(1/2) ln((1+x)/(1-x))dx =

(x/2) ln((1+x)/(1-x)) - ∫(x/2) (2/(1-x²))dx =

(x/2) ln((1+x)/(1-x)) - ∫x/(1-x²)dx =

(x/2)ln((1+x)/(1-x)) + (1/2) ln(1-x²) + C =

x arctanh(x) + (1/2) ln(1-x²) + C

B.

By parts using the formula ∫u'v = uv - ∫uv' where

u' = 1, v = arctanh(x)

u = x

v' = 1/(1-x²)

∫1 arctanh(x)dx =

x arctanh(x) - ∫x/(1-x²)dx =

x arctanh(x) + (1/2)ln(1-x²) + D