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[Hint: Put ex = t]
Let ex = t ⇒ exdx = dt
Substituting t = 1 and t = 0 in equation (1), we obtain
A = −1 and B = 1
Integrate the rational functions
Let 
Substituting x = 1, 2, and 3 respectively in equation (1), we obtain 
Integrate the rational functions
Let 
Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
−A + B = 1
−B + C = 0
On solving these equations, we obtain
From equation (1), we obtain
Integrate the rational functions
Let 
Equating the coefficients of x and constant term, we obtain
A + B = 1
2A + B = 0
On solving, we obtain
A = −1 and B = 2
Integrate the rational functions
Let 
Equating the coefficients of x and constant term, we obtain
A + B = 0
−3A + 3B = 1
On solving, we obtain
Integrate the rational functions
Let 
Substituting x = −1 and −2 in equation (1), we obtain
A = −2 and B = 4
Integrate the rational functions
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain
Let 
Substituting x = 0 and 
in equation (1), we obtain
A = 2 and B = 3
Substituting in equation (1), we obtain
Integrate the rational functions
Let
Equating the coefficients of x2, x, and constant term, we obtain
A + C = 0
−A + B = 1
−B + C = 0
On solving these equations, we obtain
From equation (1), we obtain
Integrate the rational functions
Let 
Substituting x = 1 in equation (1), we obtain
B = 4
Equating the coefficients of x2 and x, we obtain
A + C = 0
B − 2C = 3
On solving, we obtain
Integrate the rational functions
Let 
Equating the coefficients of x2 and x, we obtain
Integrate the rational functions
Let 
Substituting x = −1, −2, and 2 respectively in equation (1), we obtain
Integrate the rational functions
It can be seen that the given integrand is not a proper fraction.
Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain
Let 
Substituting x = 1 and −1 in equation (1), we obtain
Integrate the rational functions
Equating the coefficient of x2, x, and constant term, we obtain
A − B = 0
B − C = 0
A + C = 2
On solving these equations, we obtain
A = 1, B = 1, and C = 1
Integrate the rational functions
Equating the coefficient of x and constant term, we obtain
A = 3
2A + B = −1 ⇒ B = −7
Integrate the rational functions
Equating the coefficient of x3, x2, x, and constant term, we obtain
On solving these equations, we obtain
[Hint: multiply numerator and denominator by xn − 1 and put xn = t]
Multiplying numerator and denominator by xn − 1, we obtain
Substituting t = 0, −1 in equation (1), we obtain
A = 1 and B = −1
[Hint: Put sin x = t]
Substituting t = 2 and then t = 1 in equation (1), we obtain
A = 1 and B = −1
Equating the coefficients of x3, x2, x, and constant term, we obtain
A + C = 0
B + D = 4
4A + 3C = 0
4B + 3D = 10
On solving these equations, we obtain
A = 0, B = −2, C = 0, and D = 6
Let x2 = t ⇒ 2x dx = dt
Substituting t = −3 and t = −1 in equation (1), we obtain
Multiplying numerator and denominator by x3, we obtain
Let x4 = t ⇒ 4x3dx = dt
Substituting t = 0 and 1 in (1), we obtain
A = −1 and B = 1
A. 
B. 
C. 
D. 
Substituting x = 1 and 2 in (1), we obtain
A = −1 and B = 2
Hence, the correct answer is B.
A. 
B. 
C. 
D. 
Equating the coefficients of x2, x, and constant term, we obtain
A + B = 0
C = 0
A = 1
On solving these equations, we obtain
A = 1, B = −1, and C = 0
Hence, the correct answer is A.
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