Venkat tatolu has approached in a nice way ..but i think the second part of the answer is wrong ...

 log ( (1-x) + (1+x) ) dx

applying parts straightaway ..we get

 log ( (1-x) + (1+x) ) * x -  x d ( log ( (1-x) + (1+x) )) dx

now

d( log ( (1-x) + (1+x) )) is

=1/((1-x) + (1+x) ) * [ 1/ 2(1-x) *-1  + 1/2(1+x) ]

so we get

= 1/2 [((1-x) - (1+x)] / [ ((1-x) + (1+x))((1-x²) ]

multiplying and dividing by ((1-x) - (1+x)) ... we get

= 1/2 [ 2-2(1-x²)]/[ (-2x)(1-x²) ]

now x d( log ( (1-x) + (1+x) ))

thus is equal to

1/2 * x *[ 2-2(1-x²)]/[ (-2x)(1-x²) ]

= so

1/2 (1- (1-x²))/( (1-x²)) dx

1/2 sin^-1x -1/2 x

so final answer is

 log ( (1-x) + (1+x) ) * x + 1/2 sin^-1x -1/2 x +c