# RD Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.17 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 19.17 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.17 Solutions Chapter 19 Indefinite Integrals

### Question 1. ∫dx/√(2x – x2)

Solution:

We have,

Let I = ∫dx/√(2x – x2)

= ∫dx/√(1 – 1 + 2x – x2)

= ∫dx/√{1 – (x– 2x + 1)}

= ∫dx/√{1– (x – 1)2}

Let x – 1 = q …(1)

= ∫dx/√{1– (q)2}

As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So,

= sin-1(q) + C

Now put the value of q from eq(1), we get

= sin-1(x – 1) + C

### Question 2. ∫dx/√(8 + 3x – x2)

Solution:

We have,

Let I = ∫dx/√(8 + 3x – x2)

Here, (8 + 3x – x2) can be written as 8 – (x– 3x + 9/4 – 9/4)

= (8 + 9/4) – (x – 3/2)2

= (41/4) – (x – 3/2)2 Let x – 3/2 =  q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get = sin-1(2x – 3/√41) + C

### Question 3. ∫dx/√(5 – 4x – 2x2)

Solution:

We have,

Let I = ∫dx/√(5 – 4x – 2x2)

= ∫dx/√{2(5/2 – 2x – x2)}

= (1/√2)∫dx/√{5/2 – (x– 2x + 1 – 1)}

= (1/√2)∫dx/√{(5/2 + 1) – (x– 2x + 1)}

= (1/√2)∫dx/√{7/2 – (x – 1)2}

= Let x + 1 = q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get  ### Question 4. ∫dx/√(3x2 + 5x + 7)

Solution:

We have,

Let I = ∫dx/√(3x+ 5x + 7)

= ∫dx/√{3(x+ 5x/3 + 7/3)}

= (1/√3)∫dx/√{5/2-(x2-2x+1-1)}     ### Question 5. ∫dx/√{(x – α)(β – x)}

Solution:

We have,

Let I = ∫dx/√{(x – α)(β – x)}

= ∫dx/√(-x+αx + βx – αβ)

= ∫dx/√{-x+ x(α + β) – αβ}   Let x – (α + β)/2 = q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get  ### Question 6. ∫dx/√(7 – 3x – 2x2)

Solution:

We have,

Let I = ∫dx/√(7 – 3x – 2x2)

= ∫dx/√{2(7/2 – 3x/2 – x2)}    Let x + 3/2 = q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get  ### Question 7. ∫dx/√(16 – 6x – x2)

Solution:

We have,

Let I = ∫dx/√(16 – 6x – x2)

= ∫dx/√{16 – (x+ 2.3x + 9 – 9)}

= ∫dx/√{25 – (x2 + 2.3x + 9)} Let x + 3/2 = q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get Question 8. ∫dx/√(7 – 6x – x2)

Solution:

We have,

Let I = ∫dx/√(7 – 6x – x2)

= ∫dx/√{7-(x+ 2.3x + 9 – 9)}

= ∫dx/√{16 – (x+ 2.3x + 9)} Let x + 3/2 = q …(1) As we know that, ∫dx/√(a– x2) = sin-1(x/a)

So, Now put the value of q from eq(1), we get ### Question 9. ∫dx/√(5x2 – 2x)

Solution:

We have,

Let I = ∫dx/√(5x– 2x)

= ∫dx/√{5(x– 2x/5)}    I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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