Given that the antiderivative of f of x equals 1 divided by the quantity x squared plus 1 is f(x) = tan-1(x) + c, evaluate the integral from negative 1 to 1 of the 1 divided by the quantity x squared plus 1, dx . (4 points) a. pi b. pi over 2 c. pi over 4 d. 0

You know that if f(x) = 1/(x²+1) then F(x)=arctan(x).

What is arctan(x) function? It is inverse function of tan(x) in tge range (-π/2, π/2).

So if tan(x) = y then x = arctan(y)

therefore arctan(-1) = -π/4 because tan(-π/4)=-1 and arctan(1)=π/4 because tan(π/4) = 1.

In conclusion:

integral (dx/(x²+1)) from -1 to 1 =

= F(1) - F(-1) = arctan(1) - arctan(-1) =

=π/4-(-π/4)=π/4+π/4 = π/2.

answer B is correct

B. π/2.

Step-by-step explanation:

We need the value of:

1

∫ 1 / (x^2 - 1) dx  =     [ tan-1(1) - tan-1(-1) ]  = π/4 - (-π/4) = π/2.

-1

the square root of what number? if you add that, then someone will probably answer the question.

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.