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

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

3,000,540

step-by-step explanation: