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:
1 1/2 eggs
1/2 almond paste
1/8 vanilla
3/8 sugar
Step-by-step explanation:
Answer:
or around 57.88
Step-by-step explanation:
We can find the base length by using the formula for the area of a triangle.
Plug in the values that we know, the area and the height, and solve for 
.
![b=\frac{984}{17}So the base is [tex]\frac{984}{17}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7B984%7D%7B17%7D%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ESo%20the%20base%20is%20%5Btex%5D%5Cfrac%7B984%7D%7B17%7D)
or about 57.88 if you round to the nearest hundredth.
Answer:
https://demo.tds.cambiumast.com/DesmosForAssessments/TDSCalculator.html?mode=fl_scirestricted
Step-by-step explanation:
Answer:
-27
Step-by-step explanation:
-3^3 (the number after ^ is the exponent)
-3 x -3 x -3
-3 x -3 = 9 (multiplying two negatives results in a positive)
9 x -3 = -27
