NCERT Solutions for Class 12 Maths Exercise 7.5 hapter 7 Integrals – FREE PDF Download
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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.5) Exercise 7.5
Integrate the (rational) function in Exercises 1 to 6.
1. 
Ans. 
…….(i)
Comparing coefficients of 
on both sides A + B = 1 …..(ii)
Comparing constants 2A + B = 0 …..(iii)
Solving eq. (ii) and (iii), we get A = 
and B = 2
Putting these values of A and B in eq. (i),
=−log|x+1|+2log|x+2|+c=−log|x+1|+2log|x+2|+c
= 
= 
2. 
Ans. 
= 
= 
= 
3. 
Ans.
= 
…….(i)
Comparing coefficients of 
: A + B + C = 0 …….(ii)
Comparing coefficients of : –5A – 4B – 3C= 3
5A + 4B + 3C = –3 …….(iii)
Comparing constants: 6A + 3B + 2C = –1 …….(iv)
On solving eq. (i), (ii) and (iii), we get A = 1, B = –5, C = 4
Putting the values of A, B and C in eq. (i),
= 
= 
= 
4. 
Ans.
= …….(i)
Comparing coefficients of : A + B + C = 0 …….(ii)
Comparing coefficients of : –5A – 4B – 3C= 1
5A + 4B + 3C = –1 …….(iii)
Comparing constants: 6A + 3B + 2C = 0 …….(iv)
On solving eq. (i), (ii) and (iii), we get A = 
B = –2, C = 
Putting the values of A, B and C in eq. (i),
= 
= 
= 
5. 
Ans.
= 
= 
= 
= 
….(i)
Comparing coefficients of on both sides A + B = 2 …….(ii)
Comparing constants 2A + B = 0 …..(iii)
Solving eq. (ii) and (iii), we get A = 
and B = 4
Putting these values of A and B in eq. (i),
=−2log|x+1|+4log|x+2|+c=−2log|x+1|+4log|x+2|+c
=4log|x+2|−2log|x+1|+c=4log|x+2|−2log|x+1|+c
6. 
Ans.
= 
= 
= 
(Dividing numerator by denominator)
= 
…….(i)
Now 
= 
…….(ii)
Comparing coefficients of on both sides –2A + B = 
…..(iii)
Comparing constants A = 1 …….(iv)
Solving eq. (ii) and (iii), we get A = 
and B =
Putting these values of A and B in eq. (ii),
= 
= 
Putting this value in eq. (i),
= 
Integrate the following function in Exercises 7 to 12.
7. 
Ans. 
…….(i)
Comparing coefficients of 
A + C = 0 …….(ii)
Comparing coefficients of , –A + B = 1 …….(iii)
Comparing constant terms, –B + C = 0 …….(iv)
Solving eq. (ii), (iii) and (iv), we get A = 
B = 
and C =
Putting the values of A, B and C in eq. (i), 
= 
8. 
Ans.
= 
…….(i)
Comparing coefficients of : A + C = 0 …….(ii)
Comparing coefficients of : A + B – 2C= 1 …….(iii)
Comparing constants: –2A + 2B + C = 0 …….(iv)
On solving eq. (i), (ii) and (iii), we get
A = 
B = 
C = 
Putting the values of A, B and C in eq. (i), 
= 
= 
= 
= 
= 
= 
9. 
Ans.
= 
= 
= 
= 
= 
…….(i)
Comparing coefficients of : A + C = 0 …….(ii)
Comparing coefficients of : B – 2C= 3 …….(iii)
Comparing constants: –2A + B + C = 5 …….(iv)
On solving eq. (i), (ii) and (iii), we get A = B = 4, C =
Putting the values of A, B and C in eq. (i),
= 
= 
= 
= 
= 
10. 
Ans.
= 
= 
…….(i)
Comparing coefficients of : 2A + 2B + C = 0 …….(ii)
Comparing coefficients of : 5A + B = 2 …….(iii)
Comparing constants: 3A – 3B – C = –3 …….(iv)
On solving eq. (i), (ii) and (iii), we get A = 
B = 
C = 
Putting the values of A, B and C in eq. (i),
= 
= 
= 
= 
=52log|x+1|−110log|x−1|−125log|2x+3|+c=52log|x+1|−110log|x−1|−125log|2x+3|+c
11. 
Ans.
= 
= 
…….(i)
Comparing coefficients of : A + +B + C = 0 …….(ii)
Comparing coefficients of : –B + 3C= 5 …….(iii)
Comparing constants: –4A – 2B + 2C = 0 …….(iv)
On solving eq. (i), (ii) and (iii), we get A = 
B = 
C = 
Putting the values of A, B and C in eq. (i),
= 
= 
=53log|x+1|−52log|x+2|+56log|x−2|+c=53log|x+1|−52log|x+2|+56log|x−2|+c
12. 
Ans.
= 
…….(i)
[On dividing numerator by denominator]
Let 
= 
= 
…….(ii)
Comparing coefficients of : A + B = 2 …….(iii)
Comparing constants: –A + B = 1 …….(iv)
On solving eq. (iii) and (iv), we get A = B =
Putting the values of A, B and C in eq. (ii),
= 
Putting this value in eq. (i),
= 
= 
= 
Integrate the following function in Exercises 13 to 17.
13. 
Ans.
= 
…….(i)
= 
= 
Comparing the coefficients of A – B = 0 …….(ii)
Comparing the coefficients of B – C = 0 …….(iii)
Comparing constants A + C = 2 …….(iv)
On solving eq. (ii), (iii) and (iv), we get A = 1, B = 1, C = 1
Putting these values of A, B and C in eq. (i),
= 
= 
= 
= 
= 
= 
14. 
Ans. Let I = 
…….(i)
Putting 
Putting this value in eq. (i),
I = 
= 
= 
= 
= 
= 
= 
= 
= 
15. 
Ans.
= 
Putting 
,
= 
…..(i)
Comparing the coefficients of 
A + B = 0 ……(ii)
Comparing constants A – B = 1 …….(iii)
On solving the eq. (ii) and (iii), we get A = B =
Putting the values of A, B and in eq. (i),
= 
= 
16. 
Ans. Let I = 
Multiplying both numerator and denominator by 
,
I = 
= 
……..(i)
Putting 
From eq. (i),
I = 
= 
= 
= 
= 
= 
= 
= 
17. 
Ans. Let I = 
…….(i)
Putting 
From eq. (i), I = 
= 
= 
= 
= 
= 
= 
= 
= 
Integrate the following function in Exercises 18 to 21.
18. 
Ans. …….(i)
Putting 
= 
…….(ii)
Dividing numerator by denominator,
= 
….(iii)
Let 
…….(iv)
Comparing coefficients of 
A + B = –4 …….(v)
Comparing constants 4A + 3B = –10 …….(vi)
On solving eq. (v) and (vi), we get A = 2, B = –6
Putting the values of A, B and in eq. (iii),
= 
= 
= 
= 
= 
19. 
Ans. Let I = 
…….(i)
Putting 
From eq. (i),
I = 
= 
= 
= 
= 
= 
= 
= 
= 
20. 
Ans. Let I = 
= 
= 
…(i)
Putting 
Putting this value in eq. (i),
I = 
= 
I = 
= 
= 
= 
= 
= 
21. 
Ans. Let I = 
…….(i)
Putting 
From eq. (i),
I = 
= 
= 
= 
= 
= 
= 
= 
= 
= 
Choose the correct answer in each of the Exercise 22 and 23.
22. 
equals:
(A) 
(B) 
(C) 
(D) 
Ans. Let 
…….(i)
Comparing coefficients of 
A + B = 1 …….(ii)
Comparing constants –2A – B = 0 …….(iii)
On solving eq. (ii) and (iii), we get A = –1, B = 2
Putting these values of A and B in eq. (i),
= 
= 
= 
Therefore, option (B) is correct.
23. 
equals:
(A) 
(B) 
(C) 
(D) 
Ans. Let I = 
= 
= 
…(i)
Putting
Putting this value in eq. (i),
I = 
= 
I = 
= 
= 
= 
= 
= 
Therefore, option (A) is correct.