Answer:
125/6(In(x-25)) - 5/6(In(x+5))+C
Step-by-step explanation:
∫x2/x1−20x2−125dx
Should be
∫x²/(x²−20x−125)dx
First of all let's factorize the denominator.
x²−20x−125= x²+5x-25x-125
x²−20x−125= x(x+5) -25(x+5)
x²−20x−125= (x-25)(x+5)
∫x²/(x²−20x−125)dx= ∫x²/((x-25)(x+5))dx
x²/(x²−20x−125) =x²/((x-25)(x+5))
x²/((x-25)(x+5))= a/(x-25) +b/(x+5)
x²/= a(x+5) + b(x-25)
Let x=25
625 = a30
a= 625/30
a= 125/6
Let x= -5
25 = -30b
b= 25/-30
b= -5/6
x²/((x-25)(x+5))= 125/6(x-25) -5/6(x+5)
∫x²/(x²−20x−125)dx
=∫125/6(x-25) -∫5/6(x+5) Dx
= 125/6(In(x-25)) - 5/6(In(x+5))+C
131.2(477)=62,582.4 is the answer you're looking for
222222222222 my good friend
Answer:
<h2>0, 3, 8, 15</h2>
Step-by-step explanation:
Substitute n = 1, n = 2, n = 3 and n = 4 to the equation f(n) = n² - 1:
f(1) = 1² - 1 = 1 - 1 = 0
f(2) = 2² - 1 = 4 - 1 = 3
f(3) = 3² - 1 = 9 - 1 = 8
f(4) = 4² - 1 = 16 - 1 = 15
Answer:
True
Step-by-step explanation:
Both sides (right side and left side) of the equation are indeed equal.
I hope this helps!
