NCERT Solutions for Class 12 Maths Exercise 7.4 hapter 7 Integrals – FREE PDF Download
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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.4) Exercise 7.4
Integrate the following functions in Exercises 1 to 9.
1. 
Ans. Let I = 
= 
……….(i)
Putting 
From eq. (i),
I = 
= 
using ∫1a2+x2dx=1atan−1xa+c∫1a2+x2dx=1atan−1xa+c
= 
Ans.
2. 
Ans. 
= 
=
= 
3. 
Ans. 
= 
= 
= =log∣∣∣12−x+x2−4x+5√∣∣∣+c=log|12−x+x2−4x+5|+c
4. 
Ans. 
= 
= 
= 
5. 
Ans. Let I = 
= 
……….(i)
Putting 
From eq. (i),
I = 
= 
= 
= =322√tan−12–√t+c=322tan−12t+c
= =322√tan−12–√x2+c=322tan−12×2+c Ans.
6. 
Ans. Let I = 
= 
= 
……….(i)
Putting
From eq. (i),
I = 
= =13∫1(1)2−(t)2dt=13∫1(1)2−(t)2dt
= 
= 
7. 
Ans. Let I = 
= 
= 
=12∫2xx2−1√dx−log∣∣∣x+(x2−(1)2)−−−−−−−−−√∣∣∣=12∫2xx2−1dx−log|x+(x2−(1)2)|
Let I1 = 
Putting 
I1 = 
= 
= 
= 
I = 
= 
I = 
where 
8. 
Ans. Let I = 
= 
……….(i)
Putting
From eq. (i),
I = 
= 
= 
= 
9. 
Ans. Let I = 
……….(i)
Putting 
From eq. (i),
I = 
= 
= 
= 
Integrate the following functions in Exercises 10 to 18.
10. 
Ans. ∫1x2+2x+2√dx∫1×2+2x+2dx
= ∫1x2+2x+1+1√dx∫1×2+2x+1+1dx
= ∫1(x+1)2+(1)2√dx∫1(x+1)2+(1)2dx
= 
= 
11. 
Ans. 
= 
= 
= 
[For making completing the squares]
= 
= 
= 
= 
= 
= 
12. 
Ans. 
= 
= 
= 
= 
= ∫1−(x+3)2+16√dx∫1−(x+3)2+16dx
= ∫1(4)2−(x+3)2√dx∫1(4)2−(x+3)2dx
= 
13. 
Ans. 
= 
= 
= 
= 
= 
= 
14. 
Ans. 
= 
= 
= 
= 
= 
= 
= 
15. 
Ans. 
= 
= 
= 
= 
= 
= 
= 
= 
= 
16. 
Ans. Let I = 
……….(i)
Putting 
From eq. (i),
I =
=
= 
= 
= 
17. 
Ans. Let I = 
= 
= 
= 
…….(i)
Let I1 = 
= 
Putting
I1 = 
= 
= 
= 
= 
Putting this value in eq. (i),
18. 
Ans. Let I = 
……….(i)
Let numerator=Addx(denominator)+Bnumerator=Addx(denominator)+B
……(ii)
Comparing coefficients of 
6A = 5
A = 
Comparing constants,
2A + B = 
On solving, we get A = , B = 
Putting the values of A and B in eq. (ii),
Putting this value of 
in eq. (i),
I = 
I = 
I = 
……….(iii)
Now I1 = 
Putting 
I1 = 
= 
= 
……….(iv)
Again I2 =
= 
= 
= 
= 
= 
13∫1[(x+13)2+(2√3)2]dx13∫1[(x+13)2+(23)2]dx
= 
= 
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
Integrate the following functions in Exercises 19 to 23.
19. 
Ans. Let I = 
= 
……….(i)
Let Linear = A 
(Quadratic) + B
……(ii)
Comparing coefficients of 2A = 6 A = 3
Comparing constants, –9A + B = 7
On solving, we get A = 3, B = 34
Putting the values of A and B in eq. (ii),
Putting this value of 
in eq. (i),
I = 
I = 
I = 
……….(iii)
Now I1 = 
Putting 
I1 = 
=
= 
=
= 
……….(iv)
Again I2 =
= 
= 
= 
= 
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
20. 
Ans. Let I = 
……….(i)
Let Linear = A (Quadratic) + B
……(ii)
Comparing coefficients of –2A = 1 A = 
Comparing constants, 4A + B = 2
On solving, we get A = 
B = 4
Putting the values of A and B in eq. (ii),
Putting this value of 
in eq. (i),
I = 
I = 
I = 
……….(iii)
Now I1 = 
Putting 
I1 = = 
= = 
= 
…….(iv)
Again I2 =
= ∫1−x2+4x√dx∫1−x2+4xdx
= ∫1−(x2−4x)√dx∫1−(x2−4x)dx
= ∫1−(x2−4x+4−4)√dx∫1−(x2−4x+4−4)dx
= ∫1−[(x−2)2−(2)2]√dx∫1−[(x−2)2−(2)2]dx
= ∫1(2)2−(x−2)2√dx∫1(2)2−(x−2)2dx
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
21. 
Ans. Let I = 
Let Linear = A 
(Quadratic) + B
……(ii)
Comparing coefficients of 2A = 1 A = 
Comparing constants, 2A + B = 2
On solving, we get A = 
B = 1
Putting the values of A and B in eq. (ii),
Putting this value of in eq. (i),
I = 
I = 
I = 
……….(iii)
Now I1 = 
Putting 
I1 = =
= =
= 
……….(iv)
Again I2 =
= 
= 
= log∣∣∣x+1+(x+1)2+(2–√)2−−−−−−−−−−−−−−√∣∣∣log|x+1+(x+1)2+(2)2|
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
22. 
Ans. Let I = 
……….(i)
Let Linear = A (Quadratic) + B
……(ii)
Comparing coefficients of 2A = 1
A =
Comparing constants, –2A + B = 3
On solving, we get A = B = 4
Putting the values of A and B in eq. (ii), 
Putting this value of 
in eq. (i),
I = 
I = 
I = 
……….(iii)
Now I1 = 
Putting 
I1 = 
= 
= 
……….(iv)
Again I2 =
= 
= 
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
23. 
Ans. Let I = 
……….(i)
Let Linear = A (Quadratic) + B
……(ii)
Comparing coefficients of 2A = 5 A = 
Comparing constants, 4A + B = 3
On solving, we get A = B = 
Putting the values of A and B in eq. (ii),
Putting this value of 
in eq. (i),
I = 
I = 
I = 52I1−7I252I1−7I2……….(iii)
Now I1 = 
Putting 
I1 =
=
= =
= 
……….(iv)
Again I2 =
= 
= 
= log∣∣∣x+2+(x+2)2+(6–√)2−−−−−−−−−−−−−−√∣∣∣log|x+2+(x+2)2+(6)2|
= 
……….(v)
Putting values of I1 and I2 in eq. (iii),
I = 
Choose the correct answer in Exercise 24 and 25.
24. 
equals
(A) 
(B) 
(C) 
(D) 
Ans.
= 
= 
=
Therefore, option (B) is correct.
25. 
equals
(A) 
(B) 
(C) 
(D) 
Ans. Let I =
= ∫1−4x2+9x√dx∫1−4×2+9xdx
= ∫1−4(x2−94x)√dx∫1−4(x2−94x)dx
= ∫1−4(x2−94x+(98)2−(98)2)√dx∫1−4(x2−94x+(98)2−(98)2)dx
= ∫1−4[(x−98)2+(98)2]√dx∫1−4[(x−98)2+(98)2]dx
= ∫14[(98)2−(x−98)2]√dx∫14[(98)2−(x−98)2]dx
= 12∫1[(98)2−(x−98)2]√dx12∫1[(98)2−(x−98)2]dx
= 
=
Therefore, option (B) is correct.