RD Sharma Class 12 Ex 19.19 Solutions Chapter 19 Indefinite Integrals

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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. ∫((3sin⁡x – 2)cos⁡x)/(5 – cos²⁡x – 4sin⁡x) dx

Solution:

Given that I = ∫((3sin⁡x – 2)cos⁡x)/(5 – cos²x – 4sin⁡x) dx

= ∫((3sin⁡x – 2)cos⁡x)/(5 – (1 – sin²⁡x) – 4sin⁡x) dx

= ∫((3sin⁡x – 2)sin⁡x)/(5 – 1 + sin²x – 4sin⁡x) dx)

Now substitute sin⁡x = t in the above equation

cos⁡xdx = 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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