How do you evaluate the definite integral #int 5/(1+x^2)+(2x)/3dx# from #[0,1]#?

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Answer 1 · 2016-08-11T09:34:03

#1/12(15pi+4)#.

#int_0^1{5/(1+x^2)+2x/3}dx#

#=5int_0^1 1/(1+x^2)dx+2/3int_0^1xdx#

#=5[tan^-1x]_0^1+2/3[x^2/2]_0^1#

#=5[tan^-1 1-tan^-1 0]+1/3[1-0]#

#=5[pi/4-0]+1/3[1-0]#

#=5pi/4+1/3#

#=1/12(15pi+4)#.