Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.29

Question 1. âˆ«(x + 1)âˆš(x2 â€“ x + 1)dx

Solution:

We have,

âˆ«(x + 1)âˆš(x2 â€“ x + 1)dx

Let x + 1 = a d(x2 â€“ x + 1)/dx + b

=> x + 1 = a(2x â€“ 1) + b

On comparing both sides, we get,

=> 2a = 1 and b â€“ a = 1

=> a = 1/2 and b = 1 + 1/2 = 3/2

So, our equation becomes,

= âˆ«[(1/2)(2x â€“ 1) + 3/2]âˆš(x2 â€“ x + 1)dx

= (1/2)âˆ«(2x â€“ 1)âˆš(x2 â€“ x + 1)dx + (3/2)âˆ«âˆš(x2 â€“ x + 1)dx

For first part, let x2 â€“ x + 1 = t, so we have, (2x â€“ 1)dx = dt

So, we have,

= (1/2)âˆ«âˆštdt + (3/2)âˆ«âˆš[(x â€“ 1/2)2 + (âˆš3/2)2]dx

= (1/2)(2/3)(t3/2) + (3/2)âˆ«âˆš[(x â€“ 1/2)2 + (âˆš3/2)2]dx

= (1/3)(x2 â€“ x + 1)3/2 + (3/2)[(1/2)(x â€“ 1/2)âˆš(x2 â€“ x + 1) + (3/8)log|(x â€“ 1/2) + âˆš(x2 â€“ x + 1)|] + c

= (1/3)(x2 â€“ x + 1)3/2 + (3/8)(2x â€“ 1)âˆš(x2 â€“ x + 1) + (9/16) log|(x â€“ 1/2) + âˆš(x2 â€“ x + 1)| + c

Question 2. âˆ«(x + 1)âˆš(2x2 + 3)dx

Solution:

We have,

âˆ«(x + 1)âˆš(2x2 + 3)dx

Let x+1 = a d(2x2 + 3)/dx + b

=> x + 1 = a(4x) + b

On comparing both sides, we get,

=> 4a= 1 and b = 1

=> a = 1/4 and b = 1

So, the equation becomes,

= âˆ«[(1/4)(4x) + 1]âˆš(2x2 + 3)dx

= âˆ«(1/4)(4x)(x + 1)âˆš(2x2 + 3)dx + âˆ«âˆš(2x2 + 3)dx

For first part, let 2x2 + 3 = t, so we have, 4xdx = dt

So, we have,

= (1/4)âˆ«âˆštdt + âˆš2âˆ«âˆš(x2+3/2)dx

= (1/4)(2/3)(t3/2) + âˆš2[(x/2)âˆš(x2 + 3/2) + (3/4) log |x + âˆš(x2 + 3/2)|] + c

= (1/6)(2x2 + 3)3/2 + (x/2)âˆš(2x2 + 3) + (3/2âˆš2) log |x + âˆš(x2 + 3/2)| + c

Question 3. âˆ«(2x â€“ 5)âˆš(2 + 3x â€“ x2)dx

Solution:

We have,

âˆ«(2x â€“ 5)âˆš(2 + 3x â€“ x2)dx

Let 2x â€“ 5 = a d(2 + 3x â€“ x2)/dx + b

=> 2x â€“ 5 = a(3 â€“ 2x) + b

On comparing both sides, we get,

=> â€“2a= 2 and b + 3a = â€“5

=> a = â€“1 and b = â€“5 â€“ 3(â€“1) = â€“2

So, the equation becomes,

= âˆ«[(â€“1)(3 â€“ 2x) â€“ 2]âˆš(2 + 3x â€“ x2)dx

= â€“âˆ«(3 â€“ 2x)âˆš(2 + 3x â€“ x2)dx â€“ 2âˆ«âˆš(2 + 3x â€“ x2)dx

For first part, let 2 + 3x â€“ x2 = t, so we have, (3 â€“ 2x)dx = dt

So, we have,

= â€“âˆ«âˆštdt â€“ 2âˆ«âˆš[(17/4) â€“ (9/4 â€“ 3x â€“ x2)]dx

= â€“(2/3)(t3/2) â€“ 2âˆ«âˆš[(âˆš17/2)2 â€“ (x â€“ 3/2)2]dx

= â€“(2/3)(2 + 3x â€“ x2)3/2 â€“ 2[(1/2)(x â€“ 3/2)âˆš(2 + 3x â€“ x2) + (17/8) sin-1[(x â€“ 3/2)/(âˆš17/2)]] + c

= â€“(2/3)(2 + 3x â€“ x2)3/2 â€“ (1/2)(2x â€“ 3)âˆš(2 + 3x â€“ x2) â€“ (17/8) sin-1[(2x â€“ 3)/âˆš17] + c

Question 4. âˆ«(x + 2)âˆš(x2 + x + 1)dx

Solution:

We have,

âˆ«(x + 2)âˆš(x2 + x + 1)dx

Let x + 2 = a d(x2 + x + 1)/dx + b

=> x + 2 = a(2x + 1)+b

On comparing both sides, we get,

=> 2a = 1 and a + b = 2

=> a = 1/2 and b = 2 â€“ 1/2 = 3/2

So, the equation becomes,

= âˆ«[(1/2)(2x + 1) + 3/2]âˆš(x2 + x + 1)dx

= (1/2)âˆ«(2x + 1)âˆš(x2 + x + 1)dx + (3/2)âˆ«âˆš(x2 + x + 1)dx

For first part, let x2 + x + 1 = t, so we have, (2x + 1)dx = dt

So, we have,

= (1/2)âˆ«âˆštdt + (3/2)âˆ«âˆš[(x + 1/2)2 + (âˆš3/2)2]dx

=  (1/2)(2/3)(t3/2) + (3/2)[(1/2)(x + 1/2)âˆš(x2 + x + 1) + (3/8) log|(x + 1/2) + âˆš(x2 + x + 1)|] + c

= (1/3)(x2 + x + 1)3/2 + (3/8)(2x + 1)âˆš(x2 + x + 1) + (9/16) log|(x + 1/2) + âˆš(x2 + x + 1)| + c

Question 5. âˆ«(4x + 1)âˆš(x2 â€“ x â€“ 2)dx

Solution:

We have,

âˆ«(4x + 1)âˆš(x2 â€“ x â€“ 2)dx

Let 4x + 1 = a d(x2 â€“ x â€“ 2)/dx + b

=> 4x + 1 = a (2x â€“ 1) + b

Comparing both sides, we get,

=> 2a = 4 and b â€“ a = 1

=> a = 2 and b = 1 + 2

=> a = 2 and b = 3

So, the equation becomes,

= âˆ«[2(2x â€“ 1) + 3]âˆš(x2 â€“ x â€“ 2)dx

= 2âˆ«(2x â€“ 1)âˆš(x2 â€“ x â€“ 2)dx + 3âˆ«âˆš(x2 â€“ x â€“ 2)dx

For first part, let x2 â€“ x â€“ 2 = t, so we have, (2x â€“ 1)dx = dt

So, we have,

= 2âˆ«âˆštdt + 3âˆ«âˆš(x2 â€“ x â€“ 2)dx

= 2âˆ«âˆštdt + 3âˆ«âˆš[(x â€“ 1/2)2 â€“ (3/2)2]dx

= 2(2/3)(t3/2) + 3[(1/2)(x â€“ 1/2)âˆš(x2 â€“ x â€“ 2) â€“ (9/8) log|(x â€“ 1/2) + âˆš(x2 â€“ x â€“ 2)|] + c

= (4/3)(x2 â€“ x â€“ 2)3/2 + (3/4)(2x â€“ 1)âˆš(x2 â€“ x â€“ 2) â€“ (27/8) log|(x â€“ 1/2) + âˆš(x2 â€“ x â€“ 2)| + c

Question 6. âˆ«(x â€“ 2)âˆš(2x2 â€“ 6x + 5)dx

Solution:

We have,

âˆ«(x â€“ 2)âˆš(2x2 â€“ 6x + 5)dx

Let x â€“ 2 = a d(2x2 â€“ 6x + 5)/dx + b

=> x â€“ 2 = a(4x â€“ 6) + b

On comparing both sides, we get,

=> 4a = 1 and b â€“ 6a = â€“2

=> a = 1/4 and b = â€“2 + 6(1/4)

=> a = 1/4 and b = â€“1/2

So, the equation becomes,

= âˆ«[(1/4)(4x â€“ 6) + (â€“1/2)]âˆš(2x2 â€“ 6x + 5)dx

= (1/4)âˆ«(4x â€“ 6)âˆš(2x2 â€“ 6x + 5)dx â€“ (1/2)âˆ«âˆš(2x2 â€“ 6x + 5)dx

For first part, let 2x2 â€“ 6x + 5 = t, so we have, (4x â€“ 6)dx = dt

So, we have,

= (1/4)âˆ«âˆštdt â€“ (âˆš2/2)âˆ«âˆš(x2 â€“ 3x + 5/2)dx

= (1/4)(2/3)(t3/2) â€“ (âˆš2/2)âˆ«âˆš[(x â€“ 3/2)2 + (1/2)2]dx

= (1/6)(2x2 â€“ 6x + 5)3/2 â€“ (1/âˆš2)[(1/2)(x â€“ 3/2)âˆš(x2 â€“ 3x + 5/2) + (1/8) log|(x â€“ 3/2) + âˆš(x2 â€“ 3x + 5/2)|] + c

= (1/6)(2x2 â€“ 6x + 5)3/2 â€“ (1/8)(2x â€“ 3)âˆš(2x2 â€“ 6x + 5) â€“ (1/8âˆš2) log|(x â€“ 3/2) + âˆš(x2 â€“ 3x + 5/2)| + c

Question 7. âˆ«(x + 1)âˆš(x2 + x + 1)dx

Solution:

We have,

âˆ«(x + 1)âˆš(x2 + x + 1)dx

Let x + 1 = a d(x2 + x + 1)/dx + b

=> x + 1 = a(2x + 1)+b

On comparing both sides, we get,

=> 2a = 1 and a + b = 1

=> a = 1/2 and b = 1/2

So, the equation becomes,

= âˆ«[(1/2)(2x + 1) + 1/2]âˆš(x2 + x + 1)dx

= (1/2)âˆ«(2x + 1)âˆš(x2 + x + 1)dx + (1/2)âˆ«âˆš(x2 + x + 1)dx

For first part, let x2 + x + 1 = t, so we have, (2x + 1)dx = dt

So we have,

= (1/2)âˆ«âˆštdt + (1/2)âˆ«âˆš[(x + 1/2)2 + (âˆš3/2)2]dx

= (1/2)(2/3)(t3/2) + (1/2)[(1/2)(x + 1/2)âˆš(x2 + x + 1) + (3/8) log|(x + 1/2) + âˆš(x2 + x + 1)|] + c

= (1/3)(x2 + x + 1)3/2 + (1/8)(2x + 1)âˆš(x2 + x + 1) + (3/16)log|(x + 1/2) + âˆš(x2 + x + 1)| + c

Question 8. âˆ«(2x + 3)âˆš(x2 + 4x + 3)dx

Solution:

We have,

âˆ«(2x + 3)âˆš(x2 + 4x + 3)dx

Let 2x + 3 = a d(x2 + 4x + 3)/dx + b

=> 2x + 3 = a(2x + 4) + b

On comparing both sides, we get,

=> 2a = 2 and 4a + b = 3

=> a = 1 and b = 3 â€“ 4 = â€“1

So, the equation becomes,

= âˆ«[2x + 4 + (â€“1)]âˆš(x2 + 4x + 3)dx

= âˆ«(2x + 4)âˆš(x2 + 4x + 3)dx â€“ âˆ«âˆš(x2 + 4x + 3)dx

For first part, let x2 + 4x + 3 = t, so we have, (2x + 4)dx = dt

So, we have,

= âˆ«âˆštdt â€“ âˆ«âˆš(x2 + 4x + 3)dx

= (2/3)(t3/2) â€“ âˆ«âˆš[(x + 2)2â€“1]dx

= (2/3)(x2 + 4x + 3)3/2 â€“ (1/2)(x + 2)âˆš(x2 + 4x + 3) + (1/2)log |x + 2 + âˆš(x2 + 4x + 3)| + c

Question 9. âˆ«(2x â€“ 5)âˆš(x2 â€“ 4x + 3)dx

Solution:

We have,

âˆ«(2x â€“ 5)âˆš(x2 â€“ 4x + 3)dx

Let 2x â€“ 5 = a d(x2 â€“ 4x + 3)/dx + b

=> 2x â€“ 5 = a(2x â€“ 4) + b

On comparing both sides, we get,

=> 2a = 2 and b â€“ 4a = â€“5

=> a = 1 and b = â€“5 + 4(1) = â€“1

So, the equation becomes,

= âˆ«2x â€“ 4 + (â€“1)]âˆš(x2 â€“ 4x + 3)dx

= âˆ«(2x + 4)âˆš(x2 â€“ 4x + 3)dx â€“ âˆ«âˆš(x2 â€“ 4x + 3)dx

For first part, let x2 â€“ 4x + 3 = t, so we have, (2x â€“ 4)dx = dt

So, we have,

= âˆ«âˆštdt â€“ âˆ«âˆš(x2 â€“ 4x + 3)dx

= (2/3)(t3/2) â€“ âˆ«âˆš[(x â€“ 2)2 â€“ 1]dx

= (2/3)(x2 â€“ 4x + 3)3/2 â€“ (1/2)(x â€“ 2)âˆš(x2 â€“ 4x + 3) + (1/2)log |(x â€“ 2) + âˆš(x2 â€“ 4x + 3)| + c

Question 10. âˆ«xâˆš(x2 + x)dx

Solution:

We have,

âˆ«xâˆš(x2 + x)dx

Let x = a d(x2 + x)/dx + b

=> x = a(2x + 1) + b

On comparing both sides, we get,

=> 2a = 1 and a + b = 0

=> a = 1/2 and b = â€“1/2

So, the equation becomes,

= âˆ«[(1/2)(2x + 1) + (â€“1/2)]âˆš(x2 + x)dx

= (1/2)âˆ«(2x + 1)âˆš(x2 + x)dx â€“ 1/2âˆ«âˆš(x2 + x)dx

For first part, let x2 + x = t, so we have, (2x + 1)dx = dt

So, we have,

= (1/2)âˆ«âˆštdt â€“ 1/2âˆ«âˆš(x2 + x)dx

= (1/2)(2/3)(t3/2) â€“ (1/2)âˆ«âˆš[(x + 1/2)2 â€“ (1/2)2]dx

= (1/3)(t3/2) â€“ 1/2[(1/2)(x + 1/2)âˆš(x2 + x)] + (1/8)log|(x + 1/2) + âˆš(x2 + x)|] + c

= (1/3)(x2 + x)3/2 â€“ (1/8)(2x + 1)âˆš(x2 + x) + (1/16)log|(x + 1/2) + âˆš(x2 + x)| + c

Question 11. âˆ«(x â€“ 3)âˆš(x2 + 3x â€“ 18)dx

Solution:

We have,

âˆ«(x â€“ 3)âˆš(x2 + 3x â€“ 18)dx

Let x â€“ 3 = a d(x2 + 3x â€“ 18)/dx + b

=> x â€“ 3 = a(2x + 3) + b

On comparing both sides, we get,

=> 2a = 1 and 3a + b = â€“3

=> a = 1/2 and b = â€“3 â€“ 3/2 = â€“9/2

So, the equation becomes,

= âˆ«[(1/2)(2x + 3) â€“ (9/2)]âˆš(x2 + 3x â€“ 18)dx

= (1/2)âˆ«(2x + 3)âˆš(x2 + 3x â€“ 18)dx â€“ (9/2)âˆ«âˆš(x2 + 3x â€“ 18)dx

For first part, let x2 + 3x â€“ 18 = t, so we have, (2x + 3)dx = dt

So, we have,

= (1/2)âˆ«âˆštdt â€“ (9/2)âˆ«âˆš(x2 + 3x â€“ 18)dx

= (1/2)(2/3)(t3/2) â€“ (9/2)âˆ«âˆš[(x + 3/2)2 â€“ (9/2)2])dx

= (1/3)(x2 + 3x â€“ 18)3/2 â€“ (9/2)[(x + 3/2)âˆš(x2 + 3x â€“ 18) â€“ (81/8) log |(x + 3/2) + âˆš(x2 + 3x â€“ 18)|] + c

= (1/3)(x2 + 3x â€“ 18)3/2 â€“ (9/8)(2x + 3)âˆš(x2 + 3x â€“ 18) + (729/16) log |(x + 3/2) + âˆš(x2 + 3x â€“ 18)| + c

Question 12. âˆ«(x + 3)âˆš(3 â€“ 4x â€“ x2)dx

Solution:

We have,

âˆ«(x + 3)âˆš(3 â€“ 4x â€“ x2)dx

Let x + 3 = a d(3 â€“ 4x â€“ x2)/dx + b

=> x + 3 = a(â€“4 â€“ 2x) + b

On comparing both sides, we get,

=> â€“2a = 1 and b â€“ 4a = 3

=> a = â€“1/2 and b = 3 + 4(â€“1/2) = 1

So, the equation becomes,

= âˆ«[(â€“1/2)(â€“4 â€“ 2x) + 1]âˆš(3 â€“ 4x â€“ x2)dx

= (â€“1/2)âˆ«(â€“4 â€“ 2x)âˆš(3 â€“ 4x â€“ x2)dx + âˆ«âˆš(3 â€“ 4x â€“ x2)dx

For first part, let 3 â€“ 4x â€“ x2 = t, so we have, (â€“4 â€“ 2x)dx = dt

So, we have,

= (â€“1/2)âˆ«âˆštdt + âˆ«âˆš(3 â€“ 4x â€“ x2)dx

= (â€“1/2)(2/3)(t3/2) + âˆ«âˆš[(âˆš7)2 â€“ (x + 2)2]dx

= (â€“1/3)(3 â€“ 4x â€“ x2)3/2 + (1/2)[(x + 2)âˆš(3 â€“ 4x â€“ x2) + 7 tanâ€“1[(x + 2)/âˆš7]] + c

= (â€“1/3)(3 â€“ 4x â€“ x2)3/2 + (1/2)(x + 2)âˆš(3 â€“ 4x â€“ x2) + (7/2) tanâ€“1[(x + 2)/âˆš7] + c

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