Indefinite Integrals Class 12 OP Malhotra Exe-15C ISC Maths Solutions. In this article you would learn about the method to evaluate ∫dx/ax²+bx+c , i.e ∫dx/quadratic. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-12 Mathematics.

Indefinite Integrals Class 12 OP Malhotra Exe-15C ISC Maths Solutions

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-15	Indefinite Integrals
Writer	OP Malhotra
Exe-15(c)	method to evaluate ∫dx/ax²+bx+c , i.e ∫dx/quadratic

Method to Evaluate ∫dx/ax²+bx+c , i.e ∫dx / Quadratic

 Indefinite Integrals Class 12 OP Malhotra Exe-15C Solutions

Que-1: (i) ∫dx/x²+2x+10
(ii) ∫dx/(x+1)(x+2)
(iii) ∫dx/9x²+6x+1

Sol: (i) Let I = ∫dx/x²+2x+10
= ∫dx/x²+2x+1+9 = ∫dx(/x+1)²+3²
Put x + 1 = t ⇒ dx = dt
= ∫dt/t²+3² = 1/3 tan^-1 (xt/3) + C
= 1/3 tan^-1(x+1/3) + C

(ii) Let I = ∫dx/(x+1)(x+2)
= ∫dx/x²+3x+2 = ∫dx/x²+3x+9/4−1/4
= ∫dt/(x+3/2)²−(1/2)²
= 1/(2×1/2) log ∣x+3/2−1/2/x+3/2+1/2∣+C
= log ∣x+1/x+2∣+C

(iii) Let I = ∫dx/9x²+6x+1
= ∫dx/(3x+1)² = ∫(3x+1)^-2 (xdx
= (3x+1)^-2+1/(−2+1)³ + C = –1/3 1/3x+1 + C

Que-2: (i) ∫dx/9x²+6x+10
(ii) ∫dx/4x²−4x+3
(iii) ∫dx/1+x−x²

Sol: (i) Let I = ∫dx/9x²+6x+10
= 1/9∫dx/x²+2/3x+10/9
= 1/9∫dx/x²+2/3x+1/9−1/9+10/9
= 1/9∫dx/x²+2/3x+1/9−1/9+10/9
= 1/9∫dx(x+1/3)²+1²
= 1/9 × 1/1 tan^-1 (x+1/3/1) + C
= 1/9 tan^-1(3x+1/3) + C

(ii) Let I =∫dx/4x²−4x+3
= 1/4∫dx/x²−x+3/4
= 1/4∫dx/x²−x+1/4−1/4+34
= 1/4∫dx(x−1/2)²+(1/√2)²

(iii) Let I =∫dx/1+x−x² = – ∫dx/x²−x−1
= –∫dx/x²−x+1/4−5/4
= –∫dx/(x−1/2)²−(5√2)
= – 1/2×√5/2 log∣x−1/2−√5/2/x−1/2+√5/2∣ + C
= – 1/√5log∣2x−1−√5/2x−1+√5∣+C

Que-3: ∫dx/x(x⁶+1)

Sol: Let I = ∫dx/x(x⁶+1) = ∫x⁵/x⁶(x⁶+1) dx
Put x⁶ = t ⇒ 6x⁵ dx = dt
= ∫dt/6t(t+1) = 1/6∫dt/t²+t+1/4−1/4
= 16∫dt/(t+1/2)²−(1/2)²

Que-4: ∫5x−2/1+2x+3x² dx

Sol: Let I = ∫5x−2/1+2x+3x² dx
= 5/6∫(6x−12/5)/3x²+2x+1dx
= 5/6∫(6x−12/5)/3x²+2x+1dx
= 5/6∫(6x+2−2−12/5)3x²+2x+1dx
= 5/6∫6x+2/3x²+2x+1 dx + 5/6∫−22/5/3x²+2x+1 dx
= 5/6∫dt/t–11/3∫dx/3(x²+2x/3+1/3)
[Put 3x²+ 2x + 1 = t ⇒ (6x + 2)dx = dt]
= 5/6log|t| – 11/9∫dx/(x+1/3)²+(√2/3)² dx
= 5/6log|3x² + 2x + 11| – 11/9 × 3/√2 tan^-1 (x+1/3/√23) + C
= 5/6log|3x²+ 2x + 1| – 11/3√2 tan^-1(3x+1/√2) + C

Que-5: (i) ∫x/ x⁴+2x²+3 dx
(ii) ∫ e^x/e^2x+6e^x+5 dx
(iii) ∫cosx/(1−sinx)(2−sinx)dx
(iv) ∫dx/x[10+7logx+(logx)²]
(v) (3sinθ−2)cosθ/5−cos²θ−4sinθ dθ

Sol: (i) Let I = ∫x/ x⁴+2x²+3 dx
Put x² = t ⇒ 2xdx = dt
= ∫dt/2(t²+2t+3) = 1/2∫dt/t²+2t+1+2
= 1/2∫dt/(t+1)²+(√2)²
= 1/2 × 1/√2 tan^-1(t+1/√2)
[∵ ∫dx/a²+x² = 1/a tan^-1(1/a) + C]
= 1/2√2 tan^-1(x²+1/√2) + C

(ii) Let I =∫ e^x/e^2x+6e^x+5 dx
Put e^x = t ⇒ e^x dx = dt
= ∫dt/t²+6t+5 = ∫dt/t²+6t+9−4

(iii) Let I = ∫cosx/(1−sinx)(2−sinx)dx
Put sin x = t ⇒ cos x dx = dt
= ∫dt/(1−t)(2−t) = ∫dt/t²−3t+2
= ∫dt/t²−3t+9/4−9/4+2
= ∫dt/(t−3/2)²−(1/2)²
= 1/2×1/2 log ∣t−3/2−1/2/t−3/2+1/2∣ + C
[∵ ∫dx/x²−a² = 1/2a log ∣x−a/x+a∣ + C]
= log ∣t−2/t−1∣ + C = log ∣sinx−2/sinx−1∣ + C

(iv) Let I =∫dx/x[10+7logx+(logx)²]
Put log x = t ⇒ 1/x dx = dt
= ∫dt/t²+7t+10
= ∫dt/t²+7t+49/4−49/4+10
= ∫dt/(t+7/2)²−(3/2)²
= 1/2×3/2 log ∣t+7/2−3/2t+7/2+3/2∣ + C
= 1/3 log ∣t+2/t+5∣ + C
= 1/3 log ∣logx+2/logx+5∣ + C

(v) Let I =(3sinθ−2)cosθ/5−cos²θ−4sinθ dθ
= ∫(3sinθ−2)cosθ/5−(1−sin²θ)−4sinθ dθ
Put sin θ = t ⇒ cos θ dθ = dt
= ∫(3t−2)/4+t²−4t = ∫(3t−2)(t−2)² dt
= ∫3(t−2/3)/(t−2)² dt = ∫(t−2+4/3)/(t−2)² dt
= 3 ∫1/t−2 dt + 4∫dt/(t−2)²
= 3 log |t – 2| + 4 (t+1)^-2+1/-2+1 + C
= 3 log|sin θ – 2| – 4/sinθ−2 + C

–: End of Indefinite Integrals Class 12 OP Malhotra Exe-15C ISC Math Ch-15 Solution :–

Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions

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