# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.4

• Last Updated : 05 Mar, 2021

### Question 1. Integrate

Solution:

Let, I =

Use division method, then we get,

= ∫ (x + 3)dx – 4∫1/(x + 2) dx

Integrate the above equation, then we get

= x2/2 + 3x – 4 log |x + 2| + c

Hence, I = x2/2 + 3x – 4 log |x + 2| + c

### Question 2. Integrate

Solution:

Let I =

Use division method, then we get,

= ∫ x2 dx + 2∫x dx + 4∫ dx + 8 ∫1/(x – 2) dx

Integrate the above equation, then we get

= x3/3 + 2x2/2 + 4x + 8 log|x – 2| + c

= x3/3 + x2 + 4x + 8 log|x – 2| + c

Hence, I = x3/3 + x2 + 4x + 8 log|x – 2| + c

### Question 3. Integrate

Solution:

Let, I =

Use division method, then we get,

Integrate the above equation, then we get

= x2 /6 + x/9 + (43/27) log|3x + 2| + c

Hence, I = = x2 /6 + x/9 + (43/27) log|3x + 2| + c

### Question 4. Integrate

Solution:

Let, I =

We can write the above equation as below,

On solving the above equation,

= 2∫ (1/(x – 1) dx + 5 ∫(x – 1)-2 dx

Integrate the above equation, then we get

= 2 log|x – 1| + 5 (x – 1)-1/(-1) + c

= 2 log|x – 1| – 5 / (x – 1) + c

Hence, I = 2 log|x – 1| – 5 / (x – 1) + c

### Question 5. Integrate

Solution:

Let, I =

We can write the above equation as below,

=

= ∫ dx – ∫ 1/ (x + 1) dx + 2∫1/(x + 1) dx -3 ∫(x + 1)-2 dx

Integrate the above equation, then we get

= x – log|x + 1| + 2 log|x + 1| – 3(x + 1)-1/(-1) + c

= x – log|x + 1| + 2 log|x + 1| + 3/(x + 1) + c

= x + log|x + 1| + 3/(x + 1) + c

Hence, I = x + log|x + 1| + 3/(x + 1) + c

### Question 6. Integrate

Solution:

Let, I =

= 2∫ 1/(x – 1) dx + ∫(x – 1)-2 dx

Integrate the above equation, then we get

= 2 log|x – 1| + (x – 1)-1/(-1) + c

= 2 log|x – 1| – 1/(x – 1) + c

Hence, I = 2 log|x – 1| – 1/(x – 1) + c

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