Here we provide RD Sharma Class 12 Ex 19.15 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 19.15 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 | 19.15 |

Category | RD Sharma Solutions |

**RD Sharma Class 12 Ex 19.15 Solutions Chapter 19 Indefinite Integrals**

### Question 1. Evaluate ∫ 1/(4x^{2} + 12x + 5) dx

**Solution:**

Let I = ∫ 1/(4x

^{2}+ 12x + 5) dxby taking 1/4 common from the above eq

= 1/4 ∫ 1/ x

^{2}+ 3x + 5/4 dx= 1/4 ∫ 1/ x

^{2}+ 2x × (3/2)x + (3/2)^{2}– (3/2)^{2}+ 5/4 dx= 1/4 ∫ 1/ (x + 3/2)

^{2}– 1 dx (i)put (x+ 3/2) = t

dx = dt

put the above value in eq. (i)

= 1/4 ∫ 1/ t

^{2}– (1)^{2}dtIntegrate the above eq. then, we get

= 1/4 × 1/2×(1) log |t-1/t+1| +c [since, ∫1/x

^{2}– a^{2}dx = 1/2a log|x-a/x+a| +c]put the value of t in the above eq.

= 1/8 log|x+ 3/2 – 1/x+ 3/2 + 1| + c

Hence, I = 1/8 log|2x+1/ 2x+5| + c

### Question 2. Evaluate ∫1/x^{2} – 10x + 34 dx

**Solution:**

Let I = ∫1/x

^{2}– 10x + 34 dx=∫1/x

^{2}– 2x × 5 + (5)^{2}– (5)^{2}+ 34 dx=∫1/ (x – 5)

^{2}+ 9 dx (i)substituting (x-1) = t

dx = dt

put the above value in eq. (i)

= ∫ 1/ t

^{2}+ (3)^{2}dtIntegrate the above eq. then, we get

= 1/3 tan

^{-1}(t/3) + c [Since, ∫ 1/x^{2}+ a^{2}dx = 1/a tan^{-1}(x/2) + c]Put the value of t in the above eq.

Hence, I = 1/3 tan

^{-1}(x-5/ 3) + c

### Question 3. Evaluate ∫ 1/ 1-x-x^{2} dx

**Solution:**

Let I = ∫ 1/ 1-x-x

^{2}dx= ∫ 1/ -(x

^{2}– x – 1) dxadding and subtracting 1/4 in the denominator to make it a perfect square

= ∫ 1/ -(x

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

^{2}– x + 1/4] – 1 – 1/4) dx= ∫ 1/ -([x – 1/2]

^{2}– 5/4) dx= ∫ 1/ (5/4 – [x – 1/2]

^{2}) dx= ∫ 1/ ([√5/2]

^{2}– [x – 1/2]^{2}) dxIntegrate the above eq. then, we get

= 1/2(√5/2) log|√5/2 + (x-1/2)/ √5/2 – (x-1/2)| + c [since ∫ 1/a

^{2}+ x^{2}dx = 1/2a log|x+a/x-a| +c]Hence, I = 1/√5 log|√5/2 + (x-1/2) /√5/2 – (x-1/2)| + c

### Question 4. Evaluate ∫ 1/2x^{2} – x – 1 dx

**Solution:**

Let I = ∫ 1/2x

^{2}– x – 1 dxtaking 1/2 common from the above eq.

=1/2 ∫ 1/ x

^{2}– x/2 – 1/2 dx=1/2 ∫ 1/ x

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

^{2}– 9/16 dxput, x- 1/4 = t

dx = dt

= 1/2 ∫ 1/ t

^{2}– (3/4)^{2}dtIntegrate the above eq. then, we get

= (1/2) 1/[2×(3/4)] log|t-(3/4) / t+(3/4)| + c [Since, ∫ 1/x

^{2}-a^{2}dx = 1/2a log|x – a/ x+a| + c]Put the value of t in above eq.

= 1/3 log|(x-1/4-3/4)/x-1/4+3/4| + c

Hence, I = 1/3log|x – 1/2x+1| + c

### Question 5. Evaluate ∫ dx/x^{2} + 6x +13

**Solution:**

Let I =∫ dx/x

^{2}+ 6x +13 (i)We have x

^{2}+ 6x +13 = x^{2}+ 6x + 3^{2}– 3^{2}+13 =(x + 3)^{2}+ 4Put the above value in eq. (i)

∫ 1/x

^{2}+ 6x +13 dx = ∫ 1/(x+3)^{2}+ 2^{2}dxput x+3 = t and

dx = dt

= ∫ dt/ t

^{2}+ 2^{2}Integrate the above eq. then, we get

= 1/2 tan

^{-1}t/2 + cPut the value of t in above eq.

Hence, I = 1/2 tan

^{-1}x+3/2 + c

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