NCERT Solutions for Class 12 Maths Exercise 7.4 hapter 7 Integrals – FREE PDF Download
NCERT Solutions for integrals Class 12 exercises 7.4 are accessible in PDF format which is available at CoolGyan’s online learning portal. The solutions to this particular exercise from chapter 7 Integrals have been prepared by our experts who have years of experience in this field. Students can find, well-explained solutions to maths problems at CoolGyan’s online learning portal. All the solutions are provided in a way that will help you in understanding the concept effectively. You can download them in PDF format for free.
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.
