# 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.

## RD Sharma Class 12 Ex 19.4 Solutions Chapter 19 Indefinite Integrals

### 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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