# Evaluate The Integral Ln(x^2+1)dx

 Anonymous From: -Posts: -Votes: - Evaluate the integral ln(x^2+1)dx Chuck Norris From: USPosts: 28Votes: 56 Integral log(1+x^2) dx For the integrand log(x^2+1), integrate by parts, integral f dg = f g- integral g df, where f = log(x^2+1), dg = dx, df = (2 x)/(x^2+1) dx, g = x: = x log(x^2+1)- integral (2 x^2)/(x^2+1) dx Factor out constants: = x log(x^2+1)-2 integral x^2/(x^2+1) dx For the integrand x^2/(x^2+1), do long division: = x log(x^2+1)-2 integral (1-1/(x^2+1)) dx Integrate the sum term by term and factor out constants: = x log(x^2+1)+2 integral 1/(x^2+1) dx-2 integral 1 dx The integral of 1/(x^2+1) is tan^(-1)(x): = x log(x^2+1)+2 tan^(-1)(x)-2 integral 1 dx The integral of 1 is x: = x log(x^2+1)-2 x+2 tan^(-1)(x)+constant Which is equal to: = x (log(x^2+1)-2)+2 tan^(-1)(x)+constant  ComputerMonkey30 From: USPosts: 8Votes: 24 Ln(x^2 + 1)x - integral (2x^2/(x^2 + 1)) dx ln(x^2 + 1)x - 2 (x - tan^-1 (x)) + c  iBrainy From: USPosts: 7Votes: 28 Integral ln ( x ^2 + 1)dx put , u =ln( x ^2 + 1) , du = [ 1 / x ^2 + 1)]2xdx dv = dx , v = x so integral [ ln (x ^2 + 1)] dx = x ln( x ^2 + 1) - integral [ 2x^2 / ( x ^2 + 1)]dx = xln(x ^2 + 1) - 2 integral [ 1 - ( 1 / x ^2 + 1)]dx = xln(x ^2 + 1) - 2( x - arctanx) + c = x [ln(x ^2 + 1) - 2] + 2arctanx + c  SigmaPro From: USPosts: 38Votes: 114 Integration by parts. u=ln(x^2 + 1), dv= dx. du=2x/(x^2+1), v=x. Remember that der( u*v) = u dv+ v du. Integrate both sides and you get uv= int(u dv)+ int (v du). Switch it around and you get int(u dv) =uv- int (v du) So the integral is ln(x^2 + 1)*x- int(2x^2/x^2+1) Next, integrate the last part. int(2x^2/(x^2+1)). Take the 2 out for a minute. Then add a one and subtract a one in the top. int(x^2+1-1)/(x^2+1). Separate them and the integral is int(1-1/(x^2+1)). That's x - tan inverse x. Put the 2 back in, and it's 2x-2 taninvers x. Put them together and you have ln(x^2 + 1)*x-2x+2 taninverse x+c. There. Not so bad.  Enter 