How do you evaluate the definite integral #int (2x)/(1+x^2)# from #[0,1]#?

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Answer 1 · 2016-08-13T21:19:17

#int_0^1 (2x)/(1+x^2)d x=0.6931471806=l n(2)#

#int_0^1 (2x)/(1+x^2)d x=?#

#"Substitute "u=1+x^2#
#d u=2x*d x#

#int (2x)/(1+x^2)d x=int(d u)/u=l n (u)#

#"Undo substitution"#

#int_0^1 (2x)/(1+x^2)d x=|l n(1+x^2)|_0^1#

#int_0^1 (2x)/(1+x^2)d x=[l n(1+1^2]-[1+0^2]#

#int_0^1 (2x)/(1+x^2)d x=[l n (2)]-[l n(1)]#

#l n(2)=0.6931471806#
#l n(1)=0#

#int_0^1 (2x)/(1+x^2)d x=0.6931471806-0#

#int_0^1 (2x)/(1+x^2)d x=0.6931471806#