How do you find the definite integral for: #(2+3x) / (4+x2) dx# for the intervals #[0, 1]#?

Question

1422 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-23T19:01:54

# arc tan(1/2)+3/2ln(5/4).#

Suppose that, #I=int_0^1(2+3x)/(4+x^2)dx.#

#:. I=int_0^1 2/(4+x^2)dx+int_0^1(3x)/(4+x^2)dx,#

#=2int_0^1 1/(2^2+x^2)dx+3/2int_0^1 (2x)/(4+x^2)dx,#

#=2[1/2*arc tan(x/2)]_0^1+3/2int_0^1 {d/dx(4+x^2)}/(4+x^2)dx,#

#=[arc tan(1/2)-arc tan(0)]+3/2[ln(4+x^2)]_0^1,#

#=arc tan(1/2)-0+3/2[ln(4+1^2)-ln(4+0^2)],#

#=arc tan(1/2)+3/2(ln5-ln4),#

#=arc tan(1/2)+3/2ln(5/4).#