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Integration by Parts

Let u = f (x) and v = g(x) be two different functions of x. Then
 
 

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Here, choice of u and v is not arbitrary. In the product of two functions, the one that can be easily integrated should be chosen as v and the other function which is easily differentiable should be chosen as u.
 

Example
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Solution
In the given expression, the two functions can be easily integrated. But, differentiation of
x gives 1 which will make things easy. Thus, we choose u = xv = e x 
 
 
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Integration by the Method of Partial Fraction

In cases having fractions with multiple algebraic expressions in both numerator and denominator, we can simplify the given expression by writing it as addition or subtraction of fractions. This approach is known as method of partial fractions. This is applicable only when the degree of numerator is less than degree of denominator.

 

Type 1:

 

 

 

 

where A and B are constants to be determined.

 

Example
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Solution
There are multiple algebraic expressions. Degree of numerator is 1 whereas that of denominator is 2. We can express the given expression as partial fractions.
 
 
To find the values of A and B, we can substitute some simple values of x in equation (1) and find out.
When x = 0, we get A = 3/2
When x = -2, we get B = -1/2
 

 

Type 2:

 

 

 

where AB and C are constants to be determined.
 

Example
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Solution
Let us write the given expression as partial fractions.
 
 
To find the values of A and B, we can substitute some simple values of x in equation (1) and find out.
When x = -2, B = -1/5;
When x = 3, C = 6/25;
When x = 0, A = -6/25;
 

 

Type 3:

 

  

 

where A and B and C are constants to be determined.

 

Example
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Solution
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Equating the coefficient’s of x`, x and the constant terms we get A=1, B=2, C=0.
 
= log(x – 1) + log(x2 + 5) + c       [The second integral can be solved by substituting x2 + 5 = t]
= log(x – 1)(x2 + 5) + c





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