Chapter 7 Integrals Ex 7.5 Solutions
Question - 11 : - 
Answer - 11 : - 
Let 
Substituting x =−1, −2, and 2 respectively in equation (1), we obtain
Answer - 12 : -
It can be seen that thegiven 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
Question - 13 : - 
Answer - 13 : - 
Equating the coefficientof x2, x, and constant term, we obtain
A − B =0
B − C =0
A + C =2
On solving theseequations, we obtain
A =1, B = 1, and C = 1
Answer - 14 : - 
Equating the coefficientof x and constant term, we obtain
A = 3
2A + B =−1 ⇒ B =−7
Answer - 15 : - 
Equating the coefficientof x3, x2, x, andconstant term, we obtain
On solving theseequations, we obtain
[Hint: multiplynumerator and denominator by xn − 1 andput xn = t]
Answer - 16 : -
Multiplying numerator anddenominator by xn − 1, we obtain
Substituting t =0, −1 in equation (1), we obtain
A = 1and B = −1
[Hint:Put sin x = t]
Answer - 17 : - 
Substituting t =2 and then t = 1 in equation (1), we obtain
A = 1and B = −1
Answer - 18 : - 
Equating the coefficientsof x3, x2, x, andconstant term, we obtain
A + C =0
B + D =4
4A + 3C =0
4B + 3D =10
On solving theseequations, we obtain
A =0, B = −2, C = 0, and D = 6
Question - 19 : - 
Answer - 19 : -
Let x2 = t ⇒ 2x dx = dt
Substituting t =−3 and t = −1 in equation (1), we obtain
Answer - 20 : -
Multiplying numerator anddenominator by x3, we obtain
Let x4 = t ⇒ 4x3dx = dt
Substituting t =0 and 1 in (1), we obtain
A =−1 and B = 1
