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First it is necessary to notice that 4x^2-9 can be written as (2x-3)(2x+3). To solve this question, you first have to write all the fractions in terms of their lowest common denominator. In this case that...

(4-2x)/(2x+1)(x+1)(x+3) = A/(2x+1) + B/(x+1) + C/(x+3) 4-2x = A(x+1)(x+3) + B(2x+1)(x+3) + C(2x+1)(x+1) let x = -1: 4-2(-1) = B(2(-1)+1)((-1)+3) 6 = B(-1)(2) B = -3 let x = -3: 4-2(-3)= C(2(-3)+1)((-3)+1)...

A) Acceleration is the rate of change of velocity with respect to time; therefore, in order to calculate it we need to differentiate the given equation of velocity v with respect to time t: a = dv / dt = ...

*Use identity sinxcosx = 1/2sin2x

=5/2sin2x + 5cosx

*then use simple integration gives

=(-5/2 x1/2)cosx + 5sinx + c

= =5/4cosx +5sinx + c

5 + 5√ 2 - √ 8 - √ 16    (Break up all terms in terms of square roots of 2)

5 + 5√ 2 - √ 8 - 4

5 + 5√ 2 - 2 √ 2 - 4 

1 + 3√2