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.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 19 |

Exercise | 19.4 |

Category | RD Sharma Solutions |

Table of Contents

**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

= x^{2}/2 + 3x – 4 log |x + 2| + c

Hence, I = x^{2}/2 + 3x – 4 log |x + 2| + c

### Question 2. Integrate

**Solution: **

Let I =

Use division method, then we get,

= ∫ x^{2} dx + 2∫x dx + 4∫ dx + 8 ∫1/(x – 2) dx

Integrate the above equation, then we get

= x^{3}/3 + 2x^{2}/2 + 4x + 8 log|x – 2| + c

= x^{3}/3 + x^{2} + 4x + 8 log|x – 2| + c

Hence, I = x^{3}/3 + x^{2} + 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

=

= x^{2} /6 + x/9 + (43/27) log|3x + 2| + c

Hence, I = = x^{2} /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}dxIntegrate 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}dxIntegrate 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}dxIntegrate 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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