The answer is D.
Y=2x Y=x(2)-3
2=2(1) 2=3(2)-3
2=2 2=6-3
2=2
Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as
<span>(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). </span>
<span>Next, complete the square. </span>
<span>x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² </span>
<span>Also, for the x in the numerator </span>
<span>x = x + 1/2 - 1/2. </span>
<span>So </span>
<span>(x² + 2x + 3)/(x² + x + 3) = 1 + (x + 1/2)/[(x + 1/2)² + (√(11)/2)²] - 1/2/[(x + 1/2)² + (√(11)/2)²]. </span>
<span>Integrate term by term to get </span>
<span>∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C </span>
<span>b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. </span>
<span>∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C </span>
<span>= 2 √(x) ln(x) - √(x) + C. </span>
<span>c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. </span>
<span>∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C </span>
<span>= arctan(e^x) + C.</span>
Answer:
Kindly check explanation
Step-by-step explanation:
Kilogram of flour needed to complete holiday order is atleast 175kg
Number of kilograms available = 34kg
Flour comes in bags each contain 23kg of flour
He wants to buy the smallest number of bags as possible and get the amount of flour he needs.
Let F = number of bags of flours Sergei needs to buy
F = (total number of kilograms needed - number of kilograms available) / kilograms per bag
F = (≥ 175 - 34) / 23
F = ≥ 141 / 23 = ≥ 6.1304347
Since flour is purchased per bag, the smallest number of bags he can possibly buy and still get the amount of flour he need = 7 bags
Answer:
c
Step-by-step explanation:
you have to set them up as a fraction and see if they are an equivalent fraction and if they are thats the proportion
Answer:
[- 4, ∞ )
Step-by-step explanation:
the expression inside the radical must be greater than or equal to zero
x +4 ≥ 0 ⇔ x ≥ - 4
domain: x ∈ [- 4, ∞ )