Note that x² + 2x + 3 = x² + x + 3 + x. So your integrand can be written as

(x² + x + 3 + x)/(x² + x + 3) = 1 + x/(x² + x + 3). 

Next, complete the square. 

x² + x + 3 = x² + x + 1/4 + 11/4 = (x + 1/2)² + (√(11)/2)² 

Also, for the x in the numerator 

x = x + 1/2 - 1/2. 

So 

(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]. 

Integrate term by term to get 

∫ (x² + 2x + 3)/(x² + x + 3) dx = x + (1/2) ln(x² + x + 3) - (1/√(11)) arctan(2(x + 1/2)/√(11)) + C 

b) Use the fact that ln(x) = 2 ln√(x). Then put u = √(x), du = 1/[2√(x)] dx. 

∫ ln(x)/√(x) dx = 4 ∫ ln u du = 4 u ln(u) - u + C = 4√(x) ln√(x) - √(x) + C 

= 2 √(x) ln(x) - √(x) + C. 

c) There are different approaches to this. One is to multiply and divide by e^x, then use u = e^x. 

∫ 1/(e^(-x) + e^x) dx = ∫ e^x/(1 + e^(2x)) dx = ∫ du/(1 + u²) = arctan(u) + C 

= arctan(e^x) + C.

4 0

3 years ago

1.What is the value of log3 729?

xxMikexx [17]

I believe the value of log3 729 is 6