Class 12 NCERT Solutions- Mathematics Part ii – Chapter 7– Integrals Exercise 7.4

Question 1: 3x²/(x⁶ + 1)

Solution:

Let x³ = t

So, 3x²dx = dt

=> ∫3² ∕ (x⁶ + 1) dx

= ∫dt ∕ (t² + 1)

= tan^-1 t + C

= tan^-1 (x³) + C

Question 2: 1 ∕ √(1+4x²)

Solution:

Let 2x = t

So, 2dx = dt

=> ∫ 1 ∕ √(1 + 4x² )dx

= 1 ∕ 2 [ log | t + √(t² + 1) | ] + C

= 1 ∕ 2 [ log | 2x + √(4x² + 1) | ] + C

Question 3: 1 / (√(2-x)² + 1)

Solution:

Let 2 – x = t

=> -dx = dt

=> ∫ 1 ∕ ( √(2 – x)² +1) dx

= -∫ 1 ∕ (√(t² + 1) dt

= -log | t + √(t² + 1) | + C

= -log |2 – x + √((2 – x)²+ 1) | + C

Question 4: 1 ∕ √(9 – 25x²)

Solution:

Let 5x = t

So, 5dx = dt

=> ∫ 1 ∕ √ (9 – 25x²) dx

= 1 ∕ 5 ∫ 1 ∕ √ (9 – t²) dt

= 1 ∕ 5 ∫ 1 ∕ √(3² – t²) dt

= 1 ∕ 5 sin^-1(t ∕ 3) + C

= 1 ∕ 5 sin^-1(5x ∕ 3) + C

Question 5: 3x ∕ (1 + 2x⁴)

Solution:

Let √2x² = t

So, 2√2x dx = dt

∫ 3x ∕ √ (1 + 2x⁴) dx

= 3∕ 2√2 ∫ dt ∕ (1+t²)

= 3∕ 2√2 [ tan^-1 t] + C

= 3∕ 2√2 [ tan^-1(√2 x²)] + C

Question 6: x² ∕ (1 – x⁶)

Solution:

Let x³ = t

So, 3x² dx = dt

∫ x² ∕ (1 – x⁶) dx

= 1 ∕ 3 ∫ dt ∕ (1 – t²)

= 1 ∕ 3 [1 ∕ 2 log |(1 + t) ∕ (1 – t)| ] + C

= 1 ∕ 6 log |(1 + x³) ∕ (1 – x³) | + C

Question 7: (x – 1) ∕ √ (x² – 1)

Solution:

∫ (x – 1) ∕ √ (x² – 1) dx

= ∫ (x ∕ √ (x² – 1) dx – ∫ 1 ∕ √ (x² – 1) dx————- (i)

For ∫ x ∕ √ (x² – 1) dx , Let x²-1 = t => 2x dx = dt

∫ x ∕ √ (x² – 1) dx

= 1 ∕ 2 ∫ dt ∕ √t

= 1 ∕ 2 ∫ t^-1/2 dt

= 1 ∕ 2 [ 2t^1/2]

= √t

= √(x² – 1)

From (1), we obtain

∫ (x – 1) ∕ √ (x² – 1) dx

∫ (x ∕ √ (x² – 1) dx – ∫ 1 ∕ √ (x² – 1) dx

= √ (x²– 1) – log | x+ √(x²– 1)| + C

Question 8: x ∕ √ (x⁶+a⁶)

Solution:

Let x³ = t

So, 3x² dx = dt

∫ x² ∕ √(x⁶ + a⁶) dx

= ∫dt ∕ √ (t² + (a³)² )

= 1 ∕ 3 log |t + √(t² + a⁶) | + C

= 1 ∕ 3 log |x³ + √(x⁶ + a⁶) | + C

Question 9: sec²x ∕ √ (tan²x + 4)

Solution:

Let tan x = t

So, sec²x dx = dt

∫ sec²x dx ∕ √(tan²x + 4)

= ∫ dt ∕ √(t² + 2²)

= log |t + √(t² + 4) | + C

= log | tan x + √(tan²x + 4) | + C

Question 10: 1 ∕ √ (x²+2x+2)

Solution:

1 ∕ √ (x² + 2x + 2) = 1 ∕ √ ((x + 1)² + (1)²) dx

Let x + 1 = t

So, dx = dt

=> ∫ 1 ∕ √ (x² + 2x + 2) dx = ∫ 1 ∕ √(t² + 1) dt

= log | t + √(t² + 1) | + C

= log | (x + 1) + √((x + 1)² + 1) | + C

= log | (x + 1) + √(x² + 2x + 1) | + C

Question 11: 1 ∕ (9x² + 6x + 5)

Solution:

∫ 1 ∕ (9x²+6x+5) dx = ∫1 ∕ ((3x+1)² + (2)²) dx

Let (3x+1) = t

So, 3dx = dt

∫1 ∕ ((3x+1)² + (2)²) dx

= 1 ∕ 3 ∫ 1 ∕ (t²+2²) dt

= 1 ∕ 3 [ 1 ∕ 2 tan^-1❲t ∕ 2❳] + C

= 1 ∕ 6 tan^-1((3x+1) ∕ 2) + C

Question 12: 1 ∕ (√(7-6x-x²))

Solution:

7-6x-x² can be written as 7- (x²+6x+9-9)

Therefore

7- (x²+6x+9-9)

= 16 – (x²+6x+9)

= 16 – (x+3)²

= (4)² – (x+3)²

So, ∫ 1 ∕ (√(7-6x-x²) ) dx

= ∫ 1 ∕ √ ( (4)² – (x+3)² ) dx

Let x+3 = t

=> dx = dt

=> ∫ 1 ∕ √ ( (4)² – (x+3)² ) dx

=> ∫ 1 ∕ √ ( (4)² – (t)² ) dt

=> sin^-1(t ∕ 4) + C

=> sin^-1((x+3) ∕ 4) + C

Question 13: 1 ∕ √((x-1)(x-2))

Solution:

(x-1)(x-2) can be written as x²-3x+2

Therefore,

x²-3x+2

= x-3x + 9 ∕ 4 – 9 ∕ 4 + 2

= (x – 3 ∕ 2)² – 1 ∕ 4

= (x- 3 ∕ 2 )² – (1 ∕ 2)²

So, ∫ 1 ∕ √((x-3∕2)²-(1 ∕ 2)²) dx

Let x – 3 ∕ 2 = t

So, dx = dt

∫1 ∕ √((x-3∕2)²-(1 ∕ 2)²) dx

= ∫ 1 ∕ √ (t² – (1 ∕ 2)²) dt

= log | t + √(t² – (1 ∕ 2)²) + C

= log | (x- 3 ∕ 2) + √(x²-3x+2) + C

Question 14: 1 ∕ √ (8+3x-x²)

Solution:

8+3x-x² can be written as 8 – (x – 3x + 9 ∕ 4 – 9 ∕ 4)

Therefore,

8 – (x²-3x + 9 ∕ 4 – 9 ∕ 4)

= 41 ∕ 4 – (x-3 ∕ 2)²

=> ∫ 1 ∕ √ (8+3x-x²) dx = ∫ 1 ∕ √ (41 ∕ 4 – (x- 3 ∕ 2 )²) dx

Let x- 3 ∕ 2 = t

So, dx = dt

=> ∫ 1 ∕ √ (41 ∕ 4 – (x- 3 ∕ 2 )²) dx

= ∫ 1 ∕ √ ( (√41 ∕ 2)² – t²) dt

= sin^-1{t ∕ √41∕2} + C

= sin^-1{(x-3/2) ∕ √41∕2} + C

= sin^-1{(2x-3) ∕ √41} + C

Question 15: 1 ∕ √((x-a)(x-b))

Solution:

(x-a)(x-b) can be written as x² – (a+b)x +ab

Therefore,

x² – (a+b)x +ab

= x² – (a+b)x + (a+b) ∕ 4 – (a+b) ∕ 4 + ab

= [x – (a+b) ∕ 2 ]² – (a-b)² ∕ 4

= ∫ 1 ∕ √ (x-a)(x-b) dx

= ∫ 1 ∕ √{x – (a+b) ∕ 2}² – {(a-b)/2}² dx

= ∫ 1 ∕ √ [ t² – ({a-b}/2)²] dt

= log | t + √ [ t² -( {a-b}/2)²] | + C

= log | { x – (a+b)/2} + √(x-a)(x-b) | + C

Question 16: 1(4x+1) ∕ √(2x²+x-3)

Solution:

Let 4x + 1 = A d/dx (2x+x-3) + B

=> 4x +1 = A(4x+1) + B

=> 4x+1 = 4 Ax + A + B

Equating the coefficients of x and constant term on both sides, we obtain

4A = 4 => A = 1

A+B = 1 => B = 0

Let 2x² + x – 3 = t

So, (4x+1) dx = dt

=> ∫ (4x+1) ∕ √ (2x² +x-3) dx

= ∫ 1 ∕ √t dt

= 2√t + C

= 2 √2x+x-3 + C

Question 17: (x+2) ∕ √(x²-1)

Solution:

Let x+2 = A d/ dx (x²-1) + B ———– (1)

=> x + 2 = A (2x) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 1 => A = 1/2

B = 2

From (1), we obtain

(x+2) = 1/2 (2x) + 2

Then, ∫ (x+2) ∕ √ (x²-1) dx

= ∫ (1/2 (2x) + 2 ) / √ (x² – 1) dx

= 1/2 ∫ 2x ∕ √ (x² – 1) dx + ∫ 2 ∕ √(x-1) dx ———-(2)

In 1/2 ∫ 2x ∕ √ (x² – 1) dx ,

let x²-1 = t

So, 2x dx = dt

= 1/2 ∫ 2x ∕ √ (x² – 1) dx

= 1/2 ∫ dt ∕ √t

= 1/2 [2√t ]

= √t

= √(x²-1)

Then, ∫ 2 ∕ √(x²-1) dx = 2 ∫ 1 ∕ √(x²-1) dx

= 2 log | x + √ (x²-1) |

From equation (2), we obtain

∫ (x+2) ∕ √ (x²-1) dx = √ (x²-1)+ 2 log | x + √ (x²-1) | +C

Question 18: (5x-2) ∕ (1+2x+3x²)

Solution:

Let 5x-2 = A d ∕ dx (1+2x+3x²) + B

=> 5x-2 = A(2+6x) + B

Equating the coefficients of x and constant term on both sides, we obtain

5 = 6A => A = 5 ∕ 6

2A + = -2 => B = – 11 ∕ 3

So, 5x -2 = 5 ∕ 6(2+6x) + (-11 ∕ 3)

=> ∫ (5x-2) ∕ (1+2x+3x²) dx

=> ∫ {5∕6 (2+6x) – 11 ∕3} ∕ (1+2x+3x²) dx

=> 5 ∕ 6 ∫ (2+6x) ∕ (1+2x+3x²) dx – 11 ∕ 3 ∫ 1 ∕ (1+2x+3x²) dx

Let I₁ = ∫ (2+6x) ∕ (1+2x+3x²) dx and I₂ = ∫ 1 ∕ (1+2x+3x²) dx

∫ (5x-2) ∕ (1+2x+3x) dx = 5 ∕ 6 I₁ – 11 ∕ 3 I₂

I = ∫(2+6x) ∕ (1+2x+3x²) dx

Let 1+2x+3x² = t

=> (2+6x)dx = dt

I₁ = ∫ dt/ t

I₁ = log|t|

I₁ = log|1+2x+3x²|————–(2)

I₂ = ∫ 1 ∕ (1+2x+3x²) dx

1+2x+3x² can by written as 1+3{x²+2 ∕ 3 x}

Therefore,

1+3{x²+2 ∕ 3 x}

= 1+3{x² +2∕3 x + 1 ∕ 9 – 1∕ 9 }

= 1 + 3{x+1 ∕3}² – 1 ∕ 3

= 2 ∕ 3 + 3 {x + 1 ∕ 3}²

= 3 [{x+1 ∕ 3}² + 2 ∕ 9 ]

= 3 [{x+1 ∕ 3}² + {√2 ∕ 3}² ]

I = 1 ∕ 3 ∫ 1 ∕ [(x+1 ∕ 3)² + (√2 ∕ 3)² ] dx

= 1 ∕ 3 [1 ∕ (√2 / 3) tan^-1{ (x+1/3) ∕ (√2/3)}]

= 1 ∕ 3 [ 3 ∕ √2 tan^-1{ (3x+1) ∕ √2} ]

= 1 ∕ √2 tan^-1{(3x+1) ∕ √2 } ———————(3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (5x-2) ∕ (1+2x+3x²) dx = 5 ∕ 6[ log|1+2x+3×2| ] – 11/3 [ 1 ∕ √2 tan-1{(3x+1) ∕ √2 } ] + C

= 5 ∕ 6[ log|1+2x+3x²| ] – 11/3√2 tan^-1{(3x+1) ∕ √2 } + C

Question 19: (6x+7) ∕ (√((x-5)(x-4))

Solution:

(6x+7) ∕ (√((x-5)(x-4)) = (6x+7) ∕ √ (x²-9x+20)

Let 6x+7 = A d∕ dx (x²-9x+20) + B

=> 6x+7 = A(2x-9)+B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 6 => A = 3

-9A+B = 7 => B = 34

6x+7 = 3(2x-9)+34

∫ (6x+7) ∕ √(x²-9x+20) dx

= ∫ (3(2x-9)+34) ∕ √(x²-9x+20) dx

= 3 ∫ (2x-9) ∕ √(x²-9x+20) dx + 34 ∫ 1 ∕ √(x²-9x+20) dx

Let I₁ = ∫ (2x-9) ∕ √(x²-9x+20) dx and I₂ = ∫ 1 ∕ √(x²-9x+20) dx

∫ (6x+7) ∕ √(x²-9x+20) dx = 3I₁ + 34I_{2————— (1)}

I₁ = ∫ (2x-9) ∕ √(x²-9x+20) dx

Let x²-9x+20 = t

=> (2x-9)dx = dt

=> I₁ = dt / √t

I = 2√(x²-9x+20)————-(2)

and I₂ = ∫ 1 ∕ √(x²-9x+20) dx

x²-9x+20 can be written s x²-9x+20 + 81/4 – 81/4

Therefore,

x²-9x+20 + 81/4 – 81/4

= { x-9/4}² – 1/4

= { x-9/4}2 – (1/2)²

^=> I₂ = ∫ 1 ∕ √((x-9/2)² – (1/2)²) dx

I = log |{ x-9/2}+ √(x²-9x+20) |—- (3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (6x+7) ∕ √(x²-9x+20) dx

= 3[ 2√(x²-9x+20) ] + 34 log |{ x-9/2}+ √(x²-9x+20) | + C

= 6√(x²-9x+20) + 34 log |{ x-9/2}+ √(x2-9x+20) | + C

Question 20: (x+2) ∕ √(4x-x²)

Solution:

Let x+2 = A d/dx(4x-x²) + B

=> x+2 = A(4-2x) + B

Equating the coefficients of x and constant term on both sides, we obtain

-2A = 1

=> A = -1/2

4A+B = 2 => B = 4

=> (x+2) = -1/2(4-2x)+4

∫ (x+2) ∕ √(4x-x²) dx = ∫ { -1/2 (4-2x)} ∕ √(4x-x²) dx

= -1/2 ∫ (4-2x) ∕ √(4x-x²) dx + 4 ∫ 1 ∕ √(4x-x²) dx

Let I₁ = ∫ (4-2x) ∕ √(4x-x²) dx and I₂= ∫1 ∕ √(4x-x²) dx

∫ (x+2)∕ √(4x-x²) dx = -1/2 I₁ + 4I_{2—————————————-(1)}

Then, I₁ = ∫(4-2x) ∕ √(4x-x²) dx

Let 4x-x² = t

=> (4-2x)dx = dt

=> I₁ = ∫ dt∕ √(t) = 2√t = 2√(4x-x²)——————————(2)

I₂ = ∫1 ∕ √(4x-x²) dx

=> 4x-x² = -(-4x+x²)

= (-4x+x²+4-4) = 4-(x-2²)

= (2)² – (x-2)²

So, I = I1 = ∫ 1 ∕ √{(2)² – (x-2)²} dx = sin^-1{ (x-2)/2}———————(3)

Using equation (2) and (3) in equation (1), we obtain

∫ (x+2) ∕ √(4x-x²) dx = -1/2 [2√(4x-x²) ] + 4sin^-1{ (x-2)/2} + C

= -√(4x-x²) + 4sin^-1{ (x-2)/2} + C

Question 21: (x+2) ∕ √(x²+2x+3)

Solution:

∫(x+2) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ 2(x+2) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+4) ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+2) ∕ √(x²+2x+3) dx + 1 ∕ 2 ∫ 2 ∕ √(x²+2x+3) dx

1 ∕ 2 ∫ (2x+2) ∕ √(x²+2x+3) dx + ∫ 1 ∕ √(x²+2x+3) dx

Let I₁ = ∫ (2x+2) ∕ √(x²+2x+3) dx and I₂ = ∫ 1 ∕ √(x²+2x+3) dx

So, ∫(x+2) ∕ √(x²+2x+3) dx = 1 ∕ 2 I₁ + I_{2 ——————-(1)}

Then, I₁ = ∫ (2x+2) ∕ √(x²+2x+3) dx

Let x² +2x+3 = t

=> (2x+2) dx = dt

I₁ = ∫dt/ √t = 2√t = 2√(x²+2x+3)—————-(2)

I₂ = ∫ 1 ∕ √(x²+2x+3) dx

=> x²+2x+3 = x²+2x+1+2 = (x+1)² + (√2)²

I = ∫ 1 ∕ √{ (x+1)² + (√2)²} dx = log | (x+1) + √(x²+2x+3) |————– (3)

Using equation (2) and (3) in equation (1), we obtain

∫(x+2) ∕ √(x²+2x+3) dx = 1/2 [2√(x²+2x+3) ] + log | (x+1) + √(x²+2x+3) | + C

=> √(x²+2x+3)+ log | (x+1) + √(x²+2x+3) | + C

Question 22: (x+3) ∕ (x²-2x-5)

Solution:

Let (x+3) = A d/dx (x²-2x-5) + B

(x+3) = A(2x-2) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 1 => A = 1/2

-2A+B = 3 => B = 4

(x+3) = 1/2 (2x-2)+4

=> ∫ (x+3) ∕ (x²-2x-5) dx = ∫ {/2 (2x-2) + 4 } ∕ (x²-2x-5) dx

=> 1/2∫(2x-2) ∕(x²-2x-5) dx +4 ∫ 1 ∕ (x²-2x-5) dx

=> I_{1 =} ∫(2x-2) ∕ (x²-2x-5) dx and I_{2 =} ∫ 1 ∕ (x²-2x-5) dx

∫ (x+3) ∕ (x²-2x-5) dx = 1/2 I₁ + 4I₂————— (1)

Then, I₁ = ∫(2x-2) ∕ (x²-2x-5) dx

Let x²-2x-5 = t

(2x-2)dx = dt

=> I₁ = ∫ dt ∕ t = log|t| = log | x²-2x-5 |—————– (2)

I₂ = ∫ 1 ∕ (x²-2x-5) dx

= ∫ 1 ∕ {(x²-2x+1) -6} dx

= ∫ 1 ∕ {(x-1)² -(√6)²} dx

= 1 ∕ 2√6 log |{x-1-√6} ∕ {x-1+√6} |—————– (3)

Substituting equation (2) and (3) in equation (1), we obtain

∫ (x+3) ∕ (x²-2x-5) dx = 1 ∕ 2 log | x²-2x-5 | + 4 ∕ 2√6 log |{x-1-√6} ∕ {x-1+√6} | + C

= 1 ∕ 2 log | x²-2x-5 | + 2 ∕ √6 log |{x-1-√6} ∕ {x-1+√6} | + C

Question 23: (5x+3) ∕√ (x²+4x+10)

Solution:

Let 5x+3 = A d/dx (x²+4x+10) + B

=> 5x+3 = A(2x+4) + B

Equating the coefficients of x and constant term on both sides, we obtain

2A = 5 => A = 5/2

4A+B = 3 => B = -7

5X+3 = 5/2 (2X+4) – 7

∫(5x+3) ∕√ (x²+4x+10) dx

=> ∫ {5/2 (2x+4-7)} ∕ √{x²+4x+10} dx

=> 5/2 ∫(2x+) ∕ √{x²+4x+10} dx – 7 ∫1/√{x²+4x+10} dx

Let I₁ = ∫(2x+) ∕ √{x²+4x+10} dx and I₂ = ∫1/ √{x²+4x+10} dx

∫(5x+3) ∕√ (x²+4x+10) dx = 5/2 I₁ – 7I₂————-(1)

Then I₁ = ∫(2x+) ∕ √{x²+4x+10} dx

Let x²+4x+10 = t

So, (2x+4)dx = dt

=> I₁ = ∫ dt /√t = 2√(x²+4x+10) —————–(2)

I₂ = ∫1/ √{x²+4x+10} dx

= ∫1/ √{(x²+4x+4)+6} dx

= ∫1/ √{(x+2)² +(√6)²} dx

= log | (x+2)√{x²+4x+10} |——————— (3)

Using equation (2) and (3) in equation (1), we obtain

∫ (5x+3) ∕√ (x²+4x+10) dx = 5/2 [ 2√(x2+4x+10) ] -7 log | (x+2)√{x²+4x+10} | + C

= 5 [ √(x2+4x+10) ] -7 log | (x+2)√{x²+4x+10} | + C

Question 24: ∫ dx ∕ (x²+2x+2) equals

(A) x tan^-1(x+1) + C

(B) tan^-1(x+1) + C

(C) (x+1) tan^-1x + C

(D) tan^-1x + C

Solution:

To solve the integral [Tex]\int \frac{dx}{x^2+2x+2}[/Tex], we can use the method of completing the square.

First, let’s complete the square in the denominator:

[Tex]x^2 + 2x + 2 = (x^2 + 2x + 1) + 1 = (x + 1)^2 + 1[/Tex]

Now, we can rewrite the integral as:

[Tex]\int \frac{dx}{(x + 1)^2 + 1}[/Tex]

Next, we can make a substitution to simplify the integral. Let u = x + 1, then du = dx.

Substituting u = x + 1 and du = dx, we get:

[Tex]\int \frac{du}{u^2 + 1}[/Tex]

Now, this is a standard integral:

[Tex]\int \frac{du}{u^2 + 1} = \arctan(u) + C[/Tex]

Finally, substituting back u = x + 1 and adding the constant of integration C, we get the final result:

[Tex]\int \frac{dx}{x^2+2x+2} = \arctan(x + 1) + C[/Tex]

So, the solution to the integral is arctan(x + 1) + C, where C is the constant of integration.

Correct Opition is B

Question 25: ∫ dx ∕ √(9x-4x²) equals

(A) 1/9 sin^-1{ (9x-8)/8} + C

(B) 1/2 sin^-1{ (8x-9)/9} + C

(C) 1/3 sin^-1{ (9x-8)/8} + C

(D) 1/2 sin^-1{ (9x-8)/8} + C

Solution:

Factor out -4 from the square root to simplify:

[Tex]\int \frac{dx}{\sqrt{-4(x^2 – \frac{9}{4}x)}}[/Tex]

[Tex]x^2 – \frac{9}{4}x = \left(x – \frac{9}{8}\right)^2 – \left(\frac{9}{8}\right)^2[/Tex]

[Tex]= \frac{1}{2i} \int \frac{dx}{\sqrt{\left(x – \frac{9}{8}\right)^2 – \left(\frac{9}{8}\right)^2}}[/Tex]

Perform a trigonometric substitution: Let [Tex]x – \frac{9}{8} = \frac{9}{8} \sin(\theta)[/Tex]

Substitute [Tex]dx = \frac{9}{8} \cos(\theta) d\theta[/Tex] and simplify.

[Tex]x – \frac{9}{8} = \frac{9}{8} \sin(\theta)[/Tex]

[Tex]\frac{1}{2} \sin^{-1} \left( \frac{9x – 8}{8} \right) + C[/Tex]

Correct Solution: is B