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Integration of : #int(-1/2x+1/2)/(x^2 +1)dx# ?

#int(-1/2x+1/2)/(x^2 +1)dx#

1 Answer
Feb 14, 2018

Answer:

#int (-1/2x+1/2)/(x^2+1) dx = 1/2 arctanx -1/4 ln(1+x^2)+C#

Explanation:

Simplify the integrand:

#int (-1/2x+1/2)/(x^2+1) dx = 1/2 int (1-x)/(x^2+1) dx#

using the linearity of the integral:

#int (-1/2x+1/2)/(x^2+1) dx = 1/2 int dx/(1+x^2) -1/2 int (xdx)/(1+x^2)#

solve the two integrals:

#int dx/(1+x^2) = arctanx +c_1#

#int (xdx)/(1+x^2) = 1/2 int (d(1+x^2))/(1+x^2) = 1/2 ln (1+x^2) +c_2#

and substituting the partial solutions:

#int (-1/2x+1/2)/(x^2+1) dx = 1/2 arctanx -1/4 ln(1+x^2)+C#