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Description Adds a node with the value v to the list immediately after the node specified by n The node passed in n must not be null Returns a reference to the node containing the value v Adds the node passed in new to the list immediately after the node specified by n The node passed in n must not be null Throws an InvalidOperationException if n is not in the list or if new is part of another list Adds a node with the value v to the list immediately before the node specified by n The node passed in n must not be null Returns a reference to the node containing the value v Adds the node passed in new to the list immediately before the node specified by n The node passed in n must not be null Throws an InvalidOperationException if n is not in the list or if new is part of another list Adds a node with the value v to the start of the list Returns a reference to the node containing the value v Adds new to the start of the list Throws an InvalidOperationException if new is part of another list Adds a node with the value v to the end of the list Returns a reference to the node containing the value v Adds new to the end of the list Throws an InvalidOperationException if new is part of another list Returns a reference to the first node in the list that has the value v null is returned if v is not in the list Returns a reference to the last node in the list that has the value v null is returned if v is not in the list Removes the first node in the list that has the value v Returns true if the node was removed (That is, if a node with the value v was in the list and it was removed) Returns false otherwise.

public void AddAfter(LinkedListNode<T> n, LinkedListNode<T> new) public LinkedListNode<T> AddBefore(LinkedListNode<T> n, T v)

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SOLUTION In order to apply the method of partial fractions, we rst must factor the denominator of the integrand It is known that every polynomial with real coef cients will factor into linear and quadratic factors How do we nd this factorization Of course we must nd a root For a polynomial of the form x k + ak 1 x k 1 + ak 2 x k 2 + + a1 x + a0 , any integer root will be a factor of a0 This leads us to try 1, 2, 3, 6, 9 and 18 We nd that 2 and 3 are roots of x 3 4x 2 3x + 18 In point of fact, x 3 4x 2 3x + 18 = (x + 2) (x 3)2 An attempt to write A B 1 = + x+2 x 3 3x + 18 will not work We encourage the reader to try this for himself so that he will understand why an extra idea is needed In fact we will use the paradigm x3 4x 2 A 1 B C = + + x + 2 x 3 (x 3)2 x 3 4x 2 3x + 18 Putting the right-hand side over a common denominator yields A(x 3)2 + B(x + 2)(x 3) + C(x + 2) 1 = x 3 4x 2 3x + 18 x 3 4x 2 3x + 18

2 0 1 2

public void AddBefore(LinkedListNode<T> n, LinkedListNode<T> new) public LinkedList<T> AddFirst( T v) public void AddFirst(LinkedListNode new) public LinkedList<T> AddLast( T v) public void AddLast(LinkedListNode new) public LinkedList<T> Find(T v) public LinkedList<T> FindLast(T v) public bool Remove(T v)

24:

(752a)

Method public void Remove(LinkedList<T> n) public void RemoveFirst( ) public void RemoveLast( )

Of course the numerators must be equal, so 1 = A(x 3)2 + B(x + 2)(x 3) + C(x + 2) We rearrange the equation as (A + B)x 2 + ( 6A B + C)x + (9A 6B + 2C 1) = 0 Since this must be an identity in x, we arrive at the system of equations A+B =0 6A B + C =0 9A 6B + 2C 1 = 0 This system is easily solved to yield A = 1/25, B = 1/25, C = 1/5 As a result of these calculations, our integral can be transformed as follows: x3 4x 2 1 dx = 3x + 18 1/25 dx + x+2 1/25 dx + x 3 1/5 dx (x 3)2

Description Removes the node that matches n Throws an InvalidOperationException if n is not in the list Removes the first node in the list Removes the last node in the list

where Ai and Bi are the scalars de ned as Ai ri ri +1 + Bi 2 ri ri +1 s2 1 ri ri + i 2 4 1 3 ri ri + si2 2 4 (752b) (752c)

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