RD Sharma Class 12 Ex 19.4 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.4 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 19.4 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.4
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 19.4 Solutions Chapter 19 Indefinite Integrals

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

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.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Leave a Comment

Your email address will not be published.