# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.19

### Question 1. ∫ x/(x^{2} + 3x + 2) dx

**Solution:**

Given that I = ∫ x/(x

^{2 }+ 3x + 2) dxLet x = m d/dx (x

^{2 }+ 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)/(x

^{2 }+ 3x + 2) dx= 1/2 ∫(2x + 3)/(x

^{2 }+ 3x + 2) dx – 3/2 ∫1/(x^{2 }+ 3x + 2) dx= 1/2 ∫(2x + 3)/(x

^{2 }+ 3x + 2) dx – 3/2 ∫1/(x^{2 }+ 2x(3/2) + (3/2)^{2 }– (3/2)^{2 }+ 2) dx= 1/2 ∫(2x + 3)/(x

^{2 }+ 3x + 2) dx – 3/2 ∫1/((x + 3/2)^{2 }–^{ }(1/2)^{2}) dx= 1/2log|x

^{2 }+ 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 3/2 – 1/2)/(x + 3/2 + 1/2)| + cAs we know that ∫1/(a

^{2 }– x^{2})dx = 1/2a log|(x – a)/(x + a)| + c]Hence, I = 1/2log|x

^{2 }+ 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| + c

### Question 2. ∫(x + 1)/(x^{2 }+ x + 3) dx

**Solution:**

Given that I = ∫(x + 1)/(x

^{2 }+ x + 3) dxLet us considered x + 1 = m d/dx(x

^{2 }+ x + 3) + nx + 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)/(x

^{2 }+ x + 3) dx=1/2∫(2x + 1)/(x

^{2 }+ x + 1) dx + 1/2∫1/(x^{2 }+ 2x(1/2) + (1/2)²-(1/2)²+3) dx= 1/2∫(2x + 1)/(x

^{2 }+ x + 1) dx + 1/2∫1/((x + 1/2)^{2 }+ (11/4)) dx= 1/2 ∫(2x + 1)/(x

^{2 }+ x + 1) dx + 1/2 ∫1/((x + 1/2)^{2 }+ (√11/2)^{2}) dx= 1/2 log|x

^{2 }+ x + 3| + (1/2) * (1/(√11/2)) tan^{-1}((x + 1/2)/(√11/2)) + cAs we know that ∫1/(x

^{2 }+ a^{2}) dx = 1/a tan^{-1}(x/a) + cHence, I = 1/2 log|x

^{2 }+ x + 3| + 1/√11 tan^{-1}((2x + 1)/(√11)) + c

### Question 3. ∫ (x – 3)/(x^{2 }+ 2x – 4) dx

**Solution:**

Given that I = ∫(x – 3)/(x

^{2 }+ 2x – 4) dxLet us considered x – 3 = m d/dx (x

^{2 }+ 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)/(x

^{2 }+ 2x – 4) dx= 1/2 ∫(2x + 2)/(x

^{2 }+ 2x – 4) dx – 4∫ 1/(x^{2 }+ 2x + (1)^{2 }– (1)^{2 }– 4) dx= 1/2 ∫(2x + 2)/(x

^{2 }+ 2x – 4) dx – 4∫ 1/((x + 1)^{2 }– (√5)) dxAs we know that ∫1/(x

^{2 }– a^{2}) dx = 1/2a log|(x – a)/(x + a)| + c= 1/2 log|x

^{2 }+ 2x – 4| – 4 × 1/(2√5) log|(x + 1 – √5)/(x + 1 + √5)| + cHence, I = 1/2 log|x

^{2 }+ 2x – 4| – 2/√5 log|(x + 1 – √5)/(x + 1 + √5)| + c

### Question 4. ∫(2x – 3)/(x^{2 }+ 6x + 13) dx

**Solution**:

Given that I = ∫ (2x – 3)/(x

^{2 }+ 6x + 13) dxLet us considered 2x – 3 = m d/dx (x

^{2 }+ 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)/(x

^{2 }+ 6x + 13) dx= ∫(2x + 6)/(x

^{2 }+ 6x + 13) dx + ∫(-9)/(x^{2 }+ 2 * (3) * x + (3)^{2 }– (3)^{2 }+ 13) dx= ∫(2x + 6)/(x

^{2 }+ 6x + 13) dx -9 ∫1/((x + 3)^{2 }+ (2)) dx= log|(x

^{2 }+ 6x + 13)| – 9 * (1/2) tan^{-1}((x + 3)/2) + cHence, I = log|(x

^{2 }+ 6x + 13)| – 9 × (1/2) tan^{-1}((x + 3)/2) + c

### Question 5. ∫x^{2}/(x^{2 }+ 7x + 10) dx

**Solution:**

Given that I = ∫x

^{2}/(x^{2 }+ 7x + 10) dx= ∫{1 – (7x + 10)/(x

^{2 }+ 7x + 10)}dxI = x – ∫(7x + 10)/(x

^{2 }+ 7x + 10) dx + c_{1 }……..(i)Let I

_{1 }= ∫(7x + 10)/(x^{2 }+ 7x + 10) dxLet 7x + 10 = md/dx (x

^{2 }+ 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)/(3x

^{2 }– 4x + 3) dx= 1/6 ∫(6x – 4)/(3x

^{2 }– 4x + 3) dx – 1/9 ∫1/(x^{2 }– 4/3x + 1) dx= 1/6 ∫(6x – 4)/(3x

^{2 }– 4x + 3) dx – 1/9 ∫1/(x^{2 }– 2x(2/3) + (2/3)^{2 }– (2/3)^{2 }+ (2)^{2}) dx= 1/6 ∫(6x – 4)/(3x

^{2 }– 4x + 3) dx – 1/9 ∫1/(x – 2/3)^{2 }+ (√5/2)) dx= 1/6log|(3x

^{2 }– 4x + 3) | – ((1/9) × 1/(√5/3)) tan((x – 2/3)/(√5/3)) + cHence, I = 1/6log|(3x

^{2 }– 4x + 3)| – (√5/15)tan^{-1}((3x – 2)/√5) + c

### Question 6. ∫2x/(2 + x – x^{2}) dx

**Solution:**

Given that I = ∫2x/(2 + x – x

^{2}) dxNow,

2x = m(d/dx(2 + x + x

^{2})) + n2x = m(-2x + 1) + n

Now equating the co-efficient of we will get m, n

m = -1,

n = 1

∫2x/(2 + x – x

^{2}) dx= ∫(m(-2x + 1) + n)/(2 + x – x

^{2}) dx= ∫(-1(-2x + 1) + 1)/(2 + x – x

^{2}) dx= ∫(-1(-2x + 1))/(2 + x – x

^{2}) dx + 1/(2 + x – x^{2}) dx= -log|2 + x – x

^{2}| + ∫1/(2 + x – x^{2}) dx= -log|2 + x – x

^{2}| – ∫1/(x^{2 }– x – 2) dx= -log|2 + x – x

^{2}– ∫1/(x^{2 }– x(1/2)(2) + (1/2) – (1/2) – 2) dx= -log|2 + x – x

^{2}| + ∫1/((x – 1/2)^{2}– (3/2)^{2}) dx= -log|2 + x – x

^{2}| – 1/3 log|((x – 1/2) – (3/2))/((x – 1/2) + (3/2))| + cHence, I = -log|2 + x – x

^{2}| – 1/3 log|(x – 2)/(x + 1)| + c

### Question 7. ∫(1 – 3x)/(3x^{2 }+ 4x + 2) dx

**Solution:**

Given that I = ∫(1 – 3x)/(3x

^{2 }+ 4x + 2) dxLet us considered 1 – 3x = m d/dx (3x

^{2 }+ 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)/(3x

^{2 }+ 4x + 2) dx= -1/2 ∫ (6x + 4)/(3x

^{2}+ 4x + 2) dx + 3∫1/(3x^{2}+ 4x + 2) dx= -1/2 ∫ (6x + 4)/(3x

^{2}+ 4x + 2) dx + 3/3 ∫1/(x^{2}+ 4/3x + 2/3) dx= -1/2 ∫ (6x + 4)/(3x

^{2}+ 4x + 2) dx + ∫1/(x^{2}+ 2x(2/3) + (2/3)^{2 }– (2/3)^{2 }+ (2/3) dx= -1/2 ∫ (6x + 4)/(3x

^{2}+ 4x + 2) dx + ∫1/((x + 2/3)^{2}+ 2/9) dx= -1/2 ∫ (6x + 4)/(3x

^{2}+ 4x + 2) dx + ∫1/((x + 2/3)^{2}+ (√2/3)^{2}) dx= -1/2 log|(3x

^{2 }+ 4x + 2) | + 3/√2tan^{-1}((x + 2/3)/(√2/3)) + cHence, I = -1/2 log|(3x

^{2 }+ 4x + 2)| + 3/√2tan^{-1}((3x + 2)/√2) + c

### Question 8. ∫(2x + 5)/(x^{2 }– x – 2) dx

**Solution:**

Given that I = ∫(2x + 5)/(x

^{2 }– x – 2) dxLet 2x + 5 = md/dx (x

^{2 }– 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)/(x

^{2 }– x – 2) dx= ∫ ((2x – 1))/(x

^{2 }– x – 2) dx + 6∫1/(x^{2 }– 2x(1/2) + (1/2)^{2 }– (1/2)^{2 }– 2) dx= ∫ (2x – 1)/(x

^{2 }– x – 2) dx + 6∫1/((x – 1/2)^{2 }– 9/4) dx= ∫ (2x – 1)/(x

^{2 }– x – 2) dx + 6∫1/((x – 1/2)^{2 }– (3/2)^{2}) dx= log|x

^{2 }– x – 2| + 6/2(3/2) log|(x – 1/2 – 3/2)/(x – 1/2 + 3/2)| + cAs we know that ∫1/(x

^{2 }– a^{2}) dx = 1/2a log|(x – a)/(x + a)| + cHence, I = log|x

^{2 }– x – 2| + 2log|(x – 2)/(x + 1)| + c

### Question 9. ∫ (ax^{3 }+ bx)/(x^{4 }+ c^{2}) dx

**Solution:**

Given that I = ∫(ax

^{3 }+ bx)/(x^{4 }+ c^{2}) dxLet us considered ax

^{3 }+ bx = md/dx (x^{4 }+ c^{2}) + nax

^{3 }+ bx = n(4x^{3}) + nOn comparing the coefficients of x,

4m = a

m = a/4

and

n = 0

I = ∫(a/4 (4x

^{3}) + bx)/(x^{4 }+ c^{2}) dx= a/4 ∫(4x

^{3})/(x^{4 }+ c^{2}) dx + b∫x/((x^{2})^{2 }+ c^{2}) dx= a/4 ∫(4x

^{3})/(x^{4 }+ c^{2}) dx + b/2 ∫2x/((x^{2})^{2 }+ c^{2}) dx= a/4 log|x

^{4 }+ c^{2}| + b/2 I_{1 }……..(i)Now,

I

_{1 }= ∫2x/((x^{2})^{2 }+ c^{2}) dxPut x

^{2 }= t2xdx = dt

I

_{1 }= ∫1/((t)^{2 }+ c^{2}) dx= 1/c tan

^{-1}(t/c) + c_{1}l

_{1 }= 1/C tan^{-1}(x^{2}/c) + c_{1 }…………(ii)Now using equation (i) and (ii) we get,

Hence, I = a/4 log|x

^{4 }+ c^{2}| + b/2c tan(x^{2}/c) + b

### Question 10. ∫(x + 2)/(2x^{2 }+ 6x + 5) dx

**Solution:**

Given that I = ∫(x + 2)/(2x

^{2 }+ 6x + 5) dxLet us considered x + 2 = m d/dx (2x

^{2 }+ 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)/(2x

^{2 }+ 6x + 5) dx)=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/2 ∫1/(2x^{2 }+ 6x + 5) dx=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/4 ∫1/(x^{2 }+ 3x + 5/2) dx=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/4 ∫1/(x^{2 }+ 2x(3/2) + (3/2)^{2 }– (3/2)^{2 }+ 5/2) dx=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/4 ∫1/((x + 3/2)^{2 }+ 1/4) dx=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/4 ∫1/((x + 3/2)^{2}+ (1/2)^{2}) dx=1/4 ∫(4x + 6)/(2x

^{2 }+ 6x + 5) dx + 1/4 × (1/(1/2))tan^{-1}((x + 3/2)/(1/2)) + cAs we know that ∫1/(x

^{2 }+ a^{2}) dx = 1/a tan^{-1}(x/a) + cHence, I = 1/4 log|2x

^{2 }+ 6x + 5| + 1/2 tan^{-1}(2x + 3) + c

### Question 11. ∫((3sinx – 2)cosx)/(5 – cos^{2}x – 4sinx) dx

**Solution:**

Given that I = ∫((3sinx – 2)cosx)/(5 – cos

^{2}x – 4sinx) dx= ∫((3sinx – 2)cosx)/(5 – (1 – sin

^{2}x) – 4sinx) dx= ∫((3sinx – 2)sinx)/(5 – 1 + sin

^{2}x – 4sinx) dx)Now substitute sinx = t in the above equation

cosxdx = dt

So,

I = ∫(3t – 2)/(4 + t

^{2 }– 4t) dt= ∫((3t – 2))/(t

^{2}– 4t + 4) dt= ∫(3t – 2)/(t – 2)

^{2}dtNow 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)

^{2}dt= ∫(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 + 3x^{2}) dx

**Solution:**

Given that I = ∫(5x – 2)/(1 + 2x + 3x

^{2}) dxLet us considered 5x – 2 = A d/dx (1 + 2x + 3x

^{2}) + B5x – 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 + 3x

^{2}) dx becomes,I = ∫[5/6(2 + 6x) – 11/3]/(3x

^{2 }+ 2x + 1) dx= 5/6∫(2 + 6x)/(3x

^{2}+ 2x + 1) dx – 11/3∫dx/(3x^{2 }+ 2x + 1)= 5/6 log(3x

^{2}+ 2x + 1) – 11/(3 × 3)∫dx/(x^{2 }+ 2/3x + 1/3) + c= 5/6 log(3x

^{2}+ 2x + 1) – 11/9∫dx/(x^{2 }+ 2/3 x + (4/3)^{2 }+ 1/3 – (4/3)^{2}) + c= 5/6 log(3x

^{2}+ 2x + 1) – 11/9∫dx/((x + 1/3)^{2 }+(√2/3)^{2}) + c= 5/6 log(3x

^{2}+ 2x + 1) – 11/9 × 1/(√2/3) tan^{-1}(((x + 1/3)/(√2/3))] + CHence, I = 5/6 log(3x

^{2 }+ 2x + 1) – 11/(3√2) tan^{-1}[(3x + 1)/√2] + C

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