Question 1 :
Evaluate the following with respect to "x".
(x + 2) / √(x2 - 1)
Solution :
(x + 2) / √(x2 - 1) = ∫(x/√(x2 - 1)) dx + 2∫(1/√(x2 - 1)) dx
∫(x/√(x2 - 1)) dx
x2 - 1 = t
2x dx = dt
x dx = dt/2
∫(x/√(x2 - 1)) dx = ∫(1/√t) dt
= ∫t-1/2 (dt/2)
= (1/2) t1/2/(1/2)
= √(x2 - 1) ---(1)
2∫(1/√(x2 - 1)) dx = 2 log(x + √(x2 - 1)) ------(2)
(1) + (2)
= √(x2 - 1) + 2 log(x + √(x2 - 1)) + c
Question 2 :
Evaluate the following with respect to "x".
(2x + 3) / √(x2 + 4x + 1)
Solution :
∫(2x + 3) / √(x2 + 4x + 1) dx
(2x + 3) = A(d/dx) (x2 + 4x + 1) + 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 = -1
Applying the values of A and B in (1)
2x + 3 = 1 (2x + 4) - 1
By dividing each term by √(x2 + 4x + 1), we get
∫(2x + 1) / √(x2 + 4x + 1) dx
= ∫(2x+4)/√(x2 + 4x + 1) dx - 1 ∫1/√(x2 + 4x + 1)dx
= 2 log √(x2 + 4x + 1) - 1 ∫1/√(x2 + 4x + 1)dx
√(x2 + 4x + 1) = [x2 + 2x(2) + 22 - 22 + 1]
= [(x + 2)2 - 3]
= [(x + 2)2 - √32]
= 2 log √(x2 + 4x + 1) - 1 ∫1/√[(x + 2)2 - √32] dx
= 2 log √(x2 + 4x + 1) - log (x + 2) + √(x2 + 4x + 1) + c
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