Question 1 :
Evaluate the following with respect to "x".
(2x - 3) / (x2 + 4x - 12)
Solution :
∫(2x - 3) / (x2 + 4x - 12) dx
(2x - 3) = A(d/dx) (x2 + 4x - 12) + B
2x - 3 = A (2x + 4) + B ----(1)
Equating the coefficients of x.
2 = 2A
A = 1
Equating constant terms
-3 = 4A + B
-3 = 4(1) + B
-3 = 4 + B
B = -3 - 4 ===> B = -7
Applying the values of A and B in (1)
2x - 3 = 1 (2x + 4) - 7
By dividing each term by , we get
∫(2x-3)/(x2+4x-12) dx
= ∫(2x-4)/(x2+4x-12) dx - 7 ∫1/(x2+4x-12) dx
= log (x2+4x-12) - 7∫1/(x2+4x-12) dx
x2+4x-12 = x2 + 2x(2) + 22 - 22 -12
= (x + 2)2 + 4 - 12
= (x + 2)2 - 8
= (x + 2)2 - 42
= log (x2+4x-12) - 7∫1/[(x + 2)2 - 42] dx
= log (x2+4x-12) - 7 [1/2(4) log(x + 2 - 4) / (x + 2 + 4)]
= log (x2+4x-12) - (7/8) [log(x-2) / (x + 6)] + c
Question 2 :
Evaluate the following with respect to "x".
(5x - 2) / (2 + 2x + x2)
Solution :
∫(5x - 2) / (2 + 2x + x2) dx
(5x - 2) = A(d/dx) (2 + 2x + x2) + B
5x - 2 = A (2 + 2x) + B ----(1)
Equating the coefficients of x.
5 = 2A
A = 5/2
Equating constant terms
-2 = 2A + B
-2 = 2(5/2) + B
-2 = 5 + B
B = -2 - 5 ===> B = -7
Applying the values of A and B in (1)
5x - 2 = (5/2) (2 + 2x) - 7
By dividing each term by , we get
∫(5x - 2) / (2 + 2x + x2) dx
= (5/2) ∫(2x + 2) / (2 + 2x + x2) - 7 ∫1/(2 + 2x + x2) dx
= log (2 + 2x + x2) - 7∫1/(2 + 2x + x2) dx
2 + 2x + x2 = x2 + 2x(1) + 12 - 12 + 2
= (x + 1)2 - 1 + 2
= (x + 1)2 + 1
= (5/2)log (2 + 2x + x2) - 7∫1/[(x + 1)2 + 1] dx
= (5/2) log (2 + 2x + x2) - 7 tan-1 (x + 1) + c
= (5/2) log (2 + 2x + x2) - 7 tan-1 (x + 1) + c
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