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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.4
  • Last Updated : 05 Mar, 2021

Question 1. Integrate ∫ \frac{(x^2 + 5x +2)}{(x + 2)}  dx

Solution:

 Let, I = ∫ \frac{(x^2 + 5x +2)}{(x + 2)}  dx

Use division method, then we get,

\frac{(x2 + 5x +2)}{(x+2)} = x + 3 - \frac{4}{(x+2)}

∫ \frac{(x^2 + 5x +2)}{(x + 2)}  =∫ x + 3 - \frac{4}{(x+2)} dx



= ∫ (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 ∫ \frac{x^3}{(x-2)} dx

Solution: 

Let I = ∫ \frac{x^3}{(x-2)} dx

Use division method, then we get,

\frac{x^3}{(x+2)} =x^2 + 2x + 4 + \frac{8}{(x-2)}



= ∫ 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 ∫ \frac{(x^2 + x + 5)}{(3x +2)} dx

Solution:   

Let, I = ∫ \frac{(x^2 + x + 5)}{(3x +2)} dx

Use division method, then we get,

\frac{(x^2 + x + 5)}{(3x +2)} = \frac{x}{3} + \frac{1}{9} + \frac{43}{9}\times \frac{1}{3x+2} dx

∫\frac{x}{3}dx + \frac{1}{9}∫1dx + \frac{43}{9} ∫\frac{1}{(3x+2)} dx

Integrate the above equation, then we get

\frac{x^2}{6} + \frac{x}{9} + \frac{43}{9} \frac{1}{3} log|3x+2| + c

= 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 ∫ \frac{(2x+3)}{(x-1)^2} dx

Solution:

Let, I = ∫ \frac{(2x+3)}{(x-1)^2} dx

We can write the above equation as below,

∫ \frac{(2x + 2 - 2 + 3)}{(x-1)^2} dx

On solving the above equation,

∫ \frac{(2x - 2 + 5)}{(x-1)^2} dx

∫ \frac{2(x - 1)}{(x-1)^2} dx + 5 ∫\frac{1}{(x-1)^2} dx

= 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 ∫ \frac{(x^2 + 3x  - 1)}{(x+1)^2} dx

Solution:

Let, I = ∫ \frac{(x^2 + 3x  - 1)}{(x+1)^2} dx

We can write the above equation as below,

∫\frac{(x^2 + x + 2x - 1)}{(x+1)^2} dx

= ∫\frac{x(x+1)}{(x+1)^2} dx + ∫\frac{(2x - 1)}{(x+1)^2} dx

∫\frac{x}{(x+1)} dx + ∫\frac{\sqrt{2x + 2 - 2 + 1}}{(x+1)^2} dx

∫\frac{(x+1)}{(x+1)} dx - ∫\frac{1}{(x+1)} dx + 2∫\frac{(x + 1)}{(x+1)^2} dx - 3∫\frac{1}{(x+1)^2} dx

= ∫ 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 ∫ \frac{(2x - 1)}{(x - 1)^2} dx

Solution:

Let, I =∫ \frac{(2x - 1)}{(x - 1)^2} dx

∫ \frac{(2x - 1 + 2 - 2)}{(x - 1)^2} dx

∫ \frac{(2x - 2 + 1)}{(x - 1)^2} dx

∫ \frac{(2x - 2)}{(x - 1)^2} dx+ ∫\frac{1}{(x - 1)^2} dx

2∫ \frac{(x - 1)}{(x - 1)^2} dx+ ∫\frac{1}{(x - 1)^2} dx

= 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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