How do you find the definite integral for: $\frac{2 + 3 x}{4 + x 2} d x$ $(2+3x) / (4+x2) dx$ for the intervals $[0, 1]$ $[0, 1]$ ?

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Answer 1 · 2017-11-23T19:01:54

$a r c tan (\frac{1}{2}) + \frac{3}{2} ln (\frac{5}{4}) .$ $arc tan(1/2)+3/2ln(5/4).$

Suppose that, $I = \int_{0}^{1} \frac{2 + 3 x}{4 + x^{2}} d x .$ $I=int_0^1(2+3x)/(4+x^2)dx.$

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

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

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

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

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

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

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

How do you find the definite integral for: (2+3x) / (4+x2) dx2+3x4+x2dx(2+3x) / (4+x2) dx for the intervals [0, 1][0,1][0, 1]?