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

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