NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5.

- Integrals Class 12 Ex 7.1
- Integrals Class 12 Ex 7.2
- Integrals Class 12 Ex 7.3
- Integrals Class 12 Ex 7.4
- Integrals Class 12 Ex 7.6
- Integrals Class 12 Ex 7.7
- Integrals Class 12 Ex 7.8
- Integrals Class 12 Ex 7.9
- Integrals Class 12 Ex 7.10
- Integrals Class 12 Ex 7.11

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Integrals |

Exercise |
Ex 7.5 |

Number of Questions Solved |
23 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5

**Integrate the rational function in exercises 1 to 21**

**Question 1.**

**Solution:**

let ≡

⇒ x ≡ A(x+2)+B(x+1)….(i)

putting x = -1 & x = -2 in (i)

we get A = 1,B = 2

=-log|x+1| + 2log|x+2|+c

**Question 2.**

**Solution:**

let

⇒ x ≡ A(x+3)+B(x-3)…(i)

put x = 3, -3 in (i)

we get &

**Question 3.**

**Solution:**

Let

⇒ 3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(-2)…..(i)

put x = 1,2,3 in (i)

we get A = 1,B = -5 & C = 4

=log|x-1| – 5log|x-2| + 4log|x+3| + C

**Question 4.**

**Solution:**

let

⇒ x ≡ A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)…(i)

put x = 1,2,3 in (i)

**Question 5.**

**Solution:**

let

⇒ 2x = A(x+2)+B(x+1)…(i)

put x = -1, -2 in (i)

we get A = -2, B = 4

=-2log|x+1|+4log|x+2|+c

**Question 6.**

**Solution:**

is an improper fraction therefore we

convert it into a proper fraction. Divide 1 – x² by x – 2x² by long division.

**Question 7.**

**Solution:**

let

⇒ x = A(x²+1)+(Bx+C)(x-1)

Put x = 1,0

⇒

**Question 8.**

**Solution:**

⇒ x ≡ A(x-1)(x+2)+B(x+2)+C(x-1)² …(i)

put x = 1, -2

we get

**Question 9.**

**Solution:**

let

⇒ 3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)

put x = 1,-1,0

we get

**Question 10.**

**Solution:**

**Question 11.**

**Solution:**

let

**Question 12.**

**Solution:**

**Question 13.**

**Solution:**

⇒ 2 = A(1+x²) + (Bx+C)(1 -x) …(i)

Putting x = 1 in (i), we get; A = 1

Also 0 = A – B and 2 = A + C ⇒B = A = 1 & C = 1

**Question 14.**

**Solution:**

=>3x – 1 = A(x + 2) + B …(i)

Comparing coefficients A = -1 and B = -7

**Question 15.**

**Solution:**

⇒ 1 ≡ A(x-1)(x²+1) + B(x+1)(x²+1) + (Cx+D)(x+1)(x-1) ….(i)

**Question 16.**

[Hint : multiply numerator and denominator by x^{n-1} and put x^{n} = t ]

**Solution:**

**Question 17.**

**Solution:**

put sinx = t

so that cosx dx = dt

**Question 18.**

**Solution:**

put x²=y

**Question 19.**

**Solution:**

put x²=y

so that 2xdx = dy

**Question 20.**

**Solution:**

put x^{4} = t

so that 4x^{3} dx = dt

**Question 21.**

**Solution:**

Let e^{x} = t ⇒ e^{x} dx = dt

⇒

**Question 22.**

choose the correct answer in each of the following :

(a)

(b)

(c)

(d) log|(x-1)(x-2)|+c

**Solution:**

(b)

**Question 23.**

(a)

(b)

(c)

(d)

**Solution:**

(a) let

⇒ 1 = A(x²+1)+(Bx+C)(x)

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