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3 years ago

Solve c: P = a+b+c+d / 4

Mathematics

1 answer:

m_a_m_a [10]3 years ago

8 0

Answer:

as written: c = P - a - b - d/4
with parentheses: c = 4P - a - b - d

Step-by-step explanation:

The meaning of the given expression is ...

P = a + b + c + (d/4)

To solve for c, subtract all the terms on the right side not containing c.

P -(a + b + (d/4)) = c

c = P - a - b - (d/4)

_____

In such equations, parentheses are commonly missing. If that is the case here, then first we undo the division by 4, then we subtract the "not c" terms.

P = (a + b + c + d)/4 . . . . maybe what you meant

4P = a + b + c + d . . . . . . multiply by 4

4P - (a +b +d) = c

c = 4P -a -b -d

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

Also, for the x in the numerator 

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(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 

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

= arctan(e^x) + C.

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