Answer:
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C
Step-by-step explanation:
∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗
Let 1st=arctan(x)
And 2nd=1
∫▒〖arctan(x).1 dx=arctan(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗
As we know that
derivative of arctan(x)=1/(1+x^2 )
∫▒〖1 dx〗=x
So
∫▒〖arctan(x).1 dx=arctan(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1
Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now
Let 1+x^2=u
du=2xdx
Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get
1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)
1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)
1/2 ∫▒(2xdx/u) =1/2 ln(u)+C
1/2 ∫▒(2xdx/u) =1/2 ln(1+x^2 )+C
Putting values in Eq1 we get
∫▒〖arctan(x).1 dx=arctan(x).x〗-1/2 ln(1+x^2 )+C (required soultion)
(-2,0)
Lies on the X axis makes it perpendicular with the other points
The slope is 1/6 and the y intercept is 7
hope it helps
please mark as brainliest
Step-by-step explanation:
This is a probability related problem.
Probability is the likelihood of an event to occur;
Pr = 
The sample space here is from 1 to 25 which is 25
A.
Pr of a card marked 8; we have just 1 possible outcome;
Pr(8) = 
B.
Pr of drawing a card that is a multiple of 5;
Multiples of 5 = 5, 10, 15 and 25
Pr (multiples of 5) = 
C.
Pr of drawing a card with odd numbers:
Number of odd numbers between 1 and 25 = 13
Pr(odd numbers) = 
D.
Pr of drawing a number with square number on it;
Square numbers between 1 and 25 = 1, 4, 9, 16 and 25
Pr(square numbers) = 
= 
I believe 10% because 390 and 39 are exactly the same numbers almost
