Integration of Rational Function With Square Root in Denominator :
Here we are going to see some example problems to understand integration of rational function with square root in denominator
To know the formulas used in integration, please visit the page "Integration Formulas for Class 12".
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
After having gone through the stuff given above, we hope that the students would have understood, "Integration of Rational Function With Square Root in Denominator"
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