Here we provide RD Sharma Class 12 Ex 19.19 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.19 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 19 |
| Exercise | 19.19 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 19.19 Solutions Chapter 19 Indefinite Integrals
Question 1. ∫ x/(x2 + 3x + 2) dx
Solution:
Given that I = ∫ x/(x2 + 3x + 2) dx
Let x = m d/dx (x2 + 3x + 2) + n
= m(2x + 3) + n
x = (2m)x + (3λ + n)
On comparing the coefficients of x,
2m = 1
m = 1/2
3m + n = 0
3(1/2) + n = 0
n = -3/2
I = ∫(1/2(2x + 3) – 3/2)/(x2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/(x2 + 3x + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/(x2 + 2x(3/2) + (3/2)2 – (3/2)2 + 2) dx
= 1/2 ∫(2x + 3)/(x2 + 3x + 2) dx – 3/2 ∫1/((x + 3/2)2 – (1/2)2) dx
= 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 3/2 – 1/2)/(x + 3/2 + 1/2)| + c
As we know that ∫1/(a2 – x2)dx = 1/2a log|(x – a)/(x + a)| + c]
Hence, I = 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| + c
Question 2. ∫(x + 1)/(x2 + x + 3) dx
Solution:
Given that I = ∫(x + 1)/(x2 + x + 3) dx
Let us considered x + 1 = m d/dx(x2 + x + 3) + n
x + 1 = m(2x + 1) + n
x + 1 = (2m)x + (m + n)
On comparing the co-efficient of x,
2m = 1
m = 1/2
m + n = 1
(1/2) + n = 1
n = 1/2
Now,
I = ∫(1/2(2x + 1) + 1/2)/(x2 + x + 3) dx
=1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/(x2 + 2x(1/2) + (1/2)²-(1/2)²+3) dx
= 1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/((x + 1/2)2 + (11/4)) dx
= 1/2 ∫(2x + 1)/(x2 + x + 1) dx + 1/2 ∫1/((x + 1/2)2 + (√11/2)2) dx
= 1/2 log|x2 + x + 3| + (1/2) * (1/(√11/2)) tan-1((x + 1/2)/(√11/2)) + c
As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
Hence, I = 1/2 log|x2 + x + 3| + 1/√11 tan-1((2x + 1)/(√11)) + c
Question 3. ∫ (x – 3)/(x2 + 2x – 4) dx
Solution:
Given that I = ∫(x – 3)/(x2 + 2x – 4) dx
Let us considered x – 3 = m d/dx (x2 + 2x – 4) + n
= m(2x + 2) + n
x – 3 = (2m)x + (2m + n)
On comparing the coefficients of x,
2m = 1
m = 1/2
2m + n = -3
2(1/2) + n = -3
n = -4
So,
I = ∫(1/2(2x + 2) – 4)/(x2 + 2x – 4) dx
= 1/2 ∫(2x + 2)/(x2 + 2x – 4) dx – 4∫ 1/(x2 + 2x + (1)2 – (1)2 – 4) dx
= 1/2 ∫(2x + 2)/(x2 + 2x – 4) dx – 4∫ 1/((x + 1)2 – (√5)) dx
As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
= 1/2 log|x2 + 2x – 4| – 4 × 1/(2√5) log|(x + 1 – √5)/(x + 1 + √5)| + c
Hence, I = 1/2 log|x2 + 2x – 4| – 2/√5 log|(x + 1 – √5)/(x + 1 + √5)| + c
Question 4. ∫(2x – 3)/(x2 + 6x + 13) dx
Solution:
Given that I = ∫ (2x – 3)/(x2 + 6x + 13) dx
Let us considered 2x – 3 = m d/dx (x2 + 6x + 13) + n
= m(2x + 6) + n
2x – 3 = (2m)x + (6m + n)
On comparing the co-efficient of x, we get
2m = 2
m = 1
6m + n = -3
6 * 1 + n = -3
n = -9
Now,
= ∫(1 * (2x + 6) – 9)/(x2 + 6x + 13) dx
= ∫(2x + 6)/(x2 + 6x + 13) dx + ∫(-9)/(x2 + 2 * (3) * x + (3)2 – (3)2 + 13) dx
= ∫(2x + 6)/(x2 + 6x + 13) dx -9 ∫1/((x + 3)2 + (2)) dx
= log|(x2 + 6x + 13)| – 9 * (1/2) tan-1((x + 3)/2) + c
Hence, I = log|(x2 + 6x + 13)| – 9 × (1/2) tan-1((x + 3)/2) + c
Question 5. ∫x2/(x2 + 7x + 10) dx
Solution:
Given that I = ∫x2/(x2 + 7x + 10) dx
= ∫{1 – (7x + 10)/(x2 + 7x + 10)}dx
I = x – ∫(7x + 10)/(x2 + 7x + 10) dx + c1 ……..(i)
Let I1 = ∫(7x + 10)/(x2 + 7x + 10) dx
Let 7x + 10 = md/dx (x2 + 7x + 10) + n
= m(2x + 7) + n
7x + 10 = (2m)x + 7m + n
On comparing the coefficients of like powers of x,
7 = 2m
m = 7/2
7m + n = 10
7(7/2) + n = 10
n = -29/2
So, l = ∫(1/6(6x – 4) – 1/3)/(3x2 – 4x + 3) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 4/3x + 1) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 2x(2/3) + (2/3)2 – (2/3)2 + (2)2) dx
= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x – 2/3)2 + (√5/2)) dx
= 1/6log|(3x2 – 4x + 3) | – ((1/9) × 1/(√5/3)) tan((x – 2/3)/(√5/3)) + c
Hence, I = 1/6log|(3x2 – 4x + 3)| – (√5/15)tan-1((3x – 2)/√5) + c
Question 6. ∫2x/(2 + x – x2) dx
Solution:
Given that I = ∫2x/(2 + x – x2) dx
Now,
2x = m(d/dx(2 + x + x2)) + n
2x = m(-2x + 1) + n
Now equating the co-efficient of we will get m, n
m = -1,
n = 1
∫2x/(2 + x – x2) dx
= ∫(m(-2x + 1) + n)/(2 + x – x2) dx
= ∫(-1(-2x + 1) + 1)/(2 + x – x2) dx
= ∫(-1(-2x + 1))/(2 + x – x2) dx + 1/(2 + x – x2) dx
= -log|2 + x – x2| + ∫1/(2 + x – x2) dx
= -log|2 + x – x2| – ∫1/(x2 – x – 2) dx
= -log|2 + x – x2 – ∫1/(x2 – x(1/2)(2) + (1/2) – (1/2) – 2) dx
= -log|2 + x – x2| + ∫1/((x – 1/2)2 – (3/2)2) dx
= -log|2 + x – x2| – 1/3 log|((x – 1/2) – (3/2))/((x – 1/2) + (3/2))| + c
Hence, I = -log|2 + x – x2| – 1/3 log|(x – 2)/(x + 1)| + c
Question 7. ∫(1 – 3x)/(3x2 + 4x + 2) dx
Solution:
Given that I = ∫(1 – 3x)/(3x2 + 4x + 2) dx
Let us considered 1 – 3x = m d/dx (3x2 + 4x + 2) + n
= m(6x + 4) + n – 11 – 3x = (6m)× + (4λ + n)
On comparing the coefficients of l x,
6m = -3
m = -1/2
4m + n = 1
4(-1/2) + n = 1
n = 3
I = ∫(-1/2(6x + 4) + 3)/(3x2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3∫1/(3x2 + 4x + 2) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3/3 ∫1/(x2 + 4/3x + 2/3) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/(x2 + 2x(2/3) + (2/3)2 – (2/3)2 + (2/3) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + 2/9) dx
= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + (√2/3)2) dx
= -1/2 log|(3x2 + 4x + 2) | + 3/√2tan-1((x + 2/3)/(√2/3)) + c
Hence, I = -1/2 log|(3x2 + 4x + 2)| + 3/√2tan-1((3x + 2)/√2) + c
Question 8. ∫(2x + 5)/(x2 – x – 2) dx
Solution:
Given that I = ∫(2x + 5)/(x2 – x – 2) dx
Let 2x + 5 = md/dx (x2 – x – 2) + n
= m(2x – 1) + n
2x + 5 = (2m)x – m + n
On comparing the coefficients of x,
2m = 2
m = 1
-m + n = 5
-1 + n = 5
n = 6
So,
I = ∫((2x – 1) + 6)/(x2 – x – 2) dx
= ∫ ((2x – 1))/(x2 – x – 2) dx + 6∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2 – 2) dx
= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – 9/4) dx
= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – (3/2)2) dx
= log|x2 – x – 2| + 6/2(3/2) log|(x – 1/2 – 3/2)/(x – 1/2 + 3/2)| + c
As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c
Hence, I = log|x2 – x – 2| + 2log|(x – 2)/(x + 1)| + c
Question 9. ∫ (ax3 + bx)/(x4 + c2) dx
Solution:
Given that I = ∫(ax3 + bx)/(x4 + c2) dx
Let us considered ax3 + bx = md/dx (x4 + c2) + n
ax3 + bx = n(4x3) + n
On comparing the coefficients of x,
4m = a
m = a/4
and
n = 0
I = ∫(a/4 (4x3) + bx)/(x4 + c2) dx
= a/4 ∫(4x3)/(x4 + c2) dx + b∫x/((x2)2 + c2) dx
= a/4 ∫(4x3)/(x4 + c2) dx + b/2 ∫2x/((x2)2 + c2) dx
= a/4 log|x4 + c2| + b/2 I1 ……..(i)
Now,
I1 = ∫2x/((x2)2 + c2) dx
Put x2 = t
2xdx = dt
I1 = ∫1/((t)2 + c2) dx
= 1/c tan-1(t/c) + c1
l1 = 1/C tan-1(x2/c) + c1 …………(ii)
Now using equation (i) and (ii) we get,
Hence, I = a/4 log|x4 + c2| + b/2c tan(x2/c) + b
Question 10. ∫(x + 2)/(2x2 + 6x + 5) dx
Solution:
Given that I = ∫(x + 2)/(2x2 + 6x + 5) dx
Let us considered x + 2 = m d/dx (2x2 + 6x + 5) + n
= m(4x + 6) + n
x + 2 = (4m)x + (6m + n)
On comparing the coefficients of x,
So,
4m = 1
m = 1/4
6m + n = 2
6(1/4) + n = 2
n = 1/2
I = ∫(1/4(4x + 6)+1/2)/(2x2 + 6x + 5) dx)
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/2 ∫1/(2x2 + 6x + 5) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 3x + 5/2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 2x(3/2) + (3/2)2 – (3/2)2 + 5/2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + 1/4) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + (1/2)2) dx
=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 × (1/(1/2))tan-1((x + 3/2)/(1/2)) + c
As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c
Hence, I = 1/4 log|2x2 + 6x + 5| + 1/2 tan-1(2x + 3) + c
Question 11. ∫((3sinx – 2)cosx)/(5 – cos2x – 4sinx) dx
Solution:
Given that I = ∫((3sinx – 2)cosx)/(5 – cos2x – 4sinx) dx
= ∫((3sinx – 2)cosx)/(5 – (1 – sin2x) – 4sinx) dx
= ∫((3sinx – 2)sinx)/(5 – 1 + sin2x – 4sinx) dx)
Now substitute sinx = t in the above equation
cosxdx = dt
So,
I = ∫(3t – 2)/(4 + t2 – 4t) dt
= ∫((3t – 2))/(t2 – 4t + 4) dt
= ∫(3t – 2)/(t – 2)2 dt
Now Integrate partial fractions.
(3t – 2)/((t – 2)2) = A/((t – 2)) + B/((t – 2)2)
= (A(t – 2) + B)/((t – 2)2)
= (At – 2A + B)/((t – 2)2)
3t – 2 = At – 2A + B
On comparing the coefficients, we have, A = 3
and -2A + B = -2
Now, on substituting the value of A = 3 in the above equation,
-2 × 3 + B = -2
-6 + B = -2
B = 6 – 2
B = 4
So, I = ∫(3t – 2)/(t – 2)2dt
= ∫(3t – 2)/(t – 2) dt + ∫4/(t – 2)2 dt
= 3log|t – 2| – 4(1/(t – 2)) + c
= 3log|t – 2| – 4(1/(t – 2)) + c
Now put the value of t = sinx, we have ,
I = 3log|sinx – 2| – 4(1/(sinx – 2)) + c
Question 12. ∫(5x – 2)/(1 + 2x + 3x2) dx
Solution:
Given that I = ∫(5x – 2)/(1 + 2x + 3x2) dx
Let us considered 5x – 2 = A d/dx (1 + 2x + 3x2) + B
5x – 2 = A(2 + 6x) + B
5x – 2 = 6 × A + 2A + B
On comparing the Co-efficient we have, 6A = 5 and 2A + B = -2
A = 5/6
On substituting the value of A in 2A + B = -2, we n have,
2 * 5/6 + B = -2
10/6 + B = -2
B = -2 – 10/6
B = (-12 – 10)/6
B = (-22)/6
B = (-11)/3
5x – 2 = 5/6(2 + 6x) – 11/3
So, I = ∫(5x – 2)/(1 + 2x + 3x2) dx becomes,
I = ∫[5/6(2 + 6x) – 11/3]/(3x2 + 2x + 1) dx
= 5/6∫(2 + 6x)/(3x2+ 2x + 1) dx – 11/3∫dx/(3x2 + 2x + 1)
= 5/6 log(3x2 + 2x + 1) – 11/(3 × 3)∫dx/(x2 + 2/3x + 1/3) + c
= 5/6 log(3x2 + 2x + 1) – 11/9∫dx/(x2 + 2/3 x + (4/3)2 + 1/3 – (4/3)2) + c
= 5/6 log(3x2 + 2x + 1) – 11/9∫dx/((x + 1/3)2 +(√2/3)2) + c
= 5/6 log(3x2 + 2x + 1) – 11/9 × 1/(√2/3) tan-1(((x + 1/3)/(√2/3))] + C
Hence, I = 5/6 log(3x2 + 2x + 1) – 11/(3√2) tan-1[(3x + 1)/√2] + C
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