# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.17

Last Updated : 19 Apr, 2021

### Question 1. âˆ«dx/âˆš(2x – x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(2x – x2)

= âˆ«dx/âˆš(1 – 1 + 2x – x2)

= âˆ«dx/âˆš{1 – (x2 – 2x + 1)}

= âˆ«dx/âˆš{12 – (x – 1)2}

Let x – 1 = q …(1)

= âˆ«dx/âˆš{12 – (q)2}

As we know that, âˆ«dx/âˆš(a2 – 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. âˆ«dx/âˆš(8 + 3x – x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(8 + 3x – x2)

Here, (8 + 3x – x2) can be written as 8 – (x2 – 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, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

= sin-1(2x – 3/âˆš41) + C

### Question 3. âˆ«dx/âˆš(5 – 4x – 2x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(5 – 4x – 2x2)

= âˆ«dx/âˆš{2(5/2 – 2x – x2)}

= (1/âˆš2)âˆ«dx/âˆš{5/2 – (x2 – 2x + 1 – 1)}

= (1/âˆš2)âˆ«dx/âˆš{(5/2 + 1) – (x2 – 2x + 1)}

= (1/âˆš2)âˆ«dx/âˆš{7/2 – (x – 1)2}

=

Let x + 1 = q …(1)

As we know that, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

### Question 4. âˆ«dx/âˆš(3x2 + 5x + 7)

Solution:

We have,

Let I = âˆ«dx/âˆš(3x2 + 5x + 7)

= âˆ«dx/âˆš{3(x2 + 5x/3 + 7/3)}

= (1/âˆš3)âˆ«dx/âˆš{5/2-(x2-2x+1-1)}

### Question 5. âˆ«dx/âˆš{(x – Î±)(Î² – x)}

Solution:

We have,

Let I = âˆ«dx/âˆš{(x – Î±)(Î² – x)}

= âˆ«dx/âˆš(-x2 +Î±x + Î²x – Î±Î²)

= âˆ«dx/âˆš{-x2 + x(Î± + Î²) – Î±Î²}

Let x – (Î± + Î²)/2 = q …(1)

As we know that, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

### Question 6. âˆ«dx/âˆš(7 – 3x – 2x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(7 – 3x – 2x2)

= âˆ«dx/âˆš{2(7/2 – 3x/2 – x2)}

Let x + 3/2 = q …(1)

As we know that, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

### Question 7. âˆ«dx/âˆš(16 – 6x – x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(16 – 6x – x2)

= âˆ«dx/âˆš{16 – (x2 + 2.3x + 9 – 9)}

= âˆ«dx/âˆš{25 – (x2 + 2.3x + 9)}

Let x + 3/2 = q …(1)

As we know that, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

Question 8. âˆ«dx/âˆš(7 – 6x – x2)

Solution:

We have,

Let I = âˆ«dx/âˆš(7 – 6x – x2)

= âˆ«dx/âˆš{7-(x2 + 2.3x + 9 – 9)}

= âˆ«dx/âˆš{16 – (x2 + 2.3x + 9)}

Let x + 3/2 = q …(1)

As we know that, âˆ«dx/âˆš(a2 – x2) = sin-1(x/a)

So,

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

### Question 9. âˆ«dx/âˆš(5x2 – 2x)

Solution:

We have,

Let I = âˆ«dx/âˆš(5x2 – 2x)

= âˆ«dx/âˆš{5(x2 – 2x/5)}

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