Indefinite Integrals Class 12 OP Malhotra Exe-15F ISC Maths Solutions Ch-15 Solutions. In this article you would learn about the integrals of the type ∫(√(ax²+bx+c) dx i.e , ∫(√quadratic) dx . 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-15F ISC Maths Solutions Ch-15

Board	ISC
Publications	S Chand
Subject	Maths
Class	12th
Chapter-15	Indefinite Integrals
Writer	OP Malhotra
Exe-15(f)	integrals of the type ∫(√(ax²+bx+c) dx i.e , ∫(√quadratic) dx

 Integrals of the type ∫(√(ax²+bx+c) dx i.e , ∫(√quadratic) dx

 Indefinite Integrals Class 12 OP Malhotra Exe-15F Solutions

Que-1: ∫√1−x² dx

Sol: ∫√1−x² dx ;
Put x = sin θ ⇒ dx = cos θ dθ
= ∫√(1−sin²θ) cosdθ
= ∫cos²θ dθ=∫[(1+cos2θ)/2]dθ
= 1/2[θ+sin2θ/2]
= 1/2 sin^-1x + 1/2√1−x² + C
Aliter : ∫√(1²−x²) dx
= x√(1−x²)/2 + 1/2 sin^-1x/1 + C
[∵ ∫√a²−x² dx = x√a²−x² + a²/2 sin^-1(x/a) + C]

Que-2: ∫√4−x² dx

Sol: ∫√4−x² dx = ∫√2²−x² dx
= x√(4−x²)/2 + 4/2 sin^-1(x/2) + C = x√(4−x²)/2 + 2 sin^-1(x/2) + C

Que-3: ∫√3−4x² dx

Sol: ∫√3−4x² dx = 2∫√3/4−x² dx

Que-4: ∫√1+x² dx

Sol: ∫√1+x²d x = x√(1+x²)/2 + 1/2 log|x+√1+x²| + C

Que-5: ∫√16+x² dx

Sol: ∫√16+x² dx  = ∫√4²+x² dx
= x√(16+x²)/2 + 16/2 log|x + √16+x²| + C = x √(16+x²)/2 + 8 log|x + √16+x²| + C

Que-6: ∫√x²−36 dx

Sol:

Que-7: ∫√3x²+5 dx

Sol: ∫√3x²+5 dx = √3∫√x²+5/3 dx = √3∫√x²+(√5/3)²dx

Que-8: ∫x√x⁴+1 dx

Sol: I = ∫x√x⁴+1 dx;
Put x2 = t ⇒ 2xdx = dt

Que-9: ∫√1+(logx)²/x dx

Sol: Let I = ∫√1+(logx)²/x dx ;
Put log x = t ⇒ 1/x dx = dt

Que-10: ∫√x²+4x+6 dx

Sol: Let I = ∫√x²+4x+6 dx
=∫√x²+4x+4+2 dx = ∫(x+2)² + (√2)² dx
Put x + 2 = t ⇒ dx = dt
= ∫√t²+(√2)² dt = t√(t²+2)/2 + 2/2 log|t + | + C
= (x+2)√(x²+4x+6)/2 + log|x + 2 +√x²+4x+6| + C

Que-11: ∫√x²+3x dx

Sol: Let I =√x²+3x dx

Que-12: ∫√1+2x−3x² dx

Sol:

Que-13: ∫√1−4x−x² dx

Sol: Let I = ∫√1−4x−x² dx = ∫√−(x²+4x+4−5) dx = ∫√5−(x+2)² dx = ∫(√5)²−(x+2)²dx ;
Put x + 2 = t ⇒ dx = dt
= ∫√(√5)−t² dt = t√(5−t²)/2 + 5/2 sin^-1t/√5 + C
= (x+2)√(1−4x−x²)/2 + 5/2 sin^-1(x+2/√5) + C

Que-14: ∫√x(1−x) dx

Sol: Let I = ∫√x(1−x) dx = ∫√x−x² dx

Que-15: ∫√x²+4x+1 dx

Sol: Let I =∫√x²+4x+1 dx
= ∫√x²+4x+4-3 dx= ∫√(x+2)²−(√3)² dx
= (x+2)√((x+2)²−3)/2 – 3/2 log|x + 2 + √((x+2)²−3)| + C
[∵ ∫√x²−a² dx = x√(x²−a²)/2 – a²/2 log|x +√x²−a²| + C]
= (x+2)√(x²+4x+1)/2 – log|x + 2 +| + C

Que-16: ∫(2x+3)√x²+4x+3 dx

Sol: Let I =∫(2x+3)√x²+4x+3 dx
= ∫(2x+4−1)√x²+4x+3 dx
= ∫(2x+4)√x²+4x+3 dx– ∫√x²+4x+3 dx = ∫√tdt – ∫√(x²+4x+4-1) dx
[Put x² + 4x + 3 = t ⇒ (2x + 4) dx = dt]
= t^1/2+1/1/2+1 – ∫√(x+2)²−1² dx
= 2/3 t^3/2– [(x+2)√((x+2)²−1)/2 – 1/2 log | x + 2 + √(x+2)²−1|] + C
= 2/3 (x²+4x+3)^3/2 – (x+2/2) √x²+4x+3 + 1/2 log|x + 2 +√(x+2)²−1| + C

Que-17: ∫(2x−5)√x²−4x+3 dx

Sol: Let I =∫(2x−5)√x²−4x+3 dx
= ∫(2x−4−1)√(x²−4x+3) dx = ∫√x²−4x+3(2x−4)dx – ∫√x²−4x+3 dx
Put x² + 4x + 3 = t ⇒ (2x – 4) dx = dt

Que-18: ∫x√x+x² dx

Sol: Let I = ∫x√x+x² dx = 1/2∫(2x+1−1)√(x+x²) dx = 1/2∫(x+x²)^1/2 (2x+1) dx – 1/2∫√x²+x cdx
Put x + x² = t ⇒ (1 + 2x) dx = dt

Que-19: ∫(x−5)√x²+x dx

Sol: Let I =∫(x−5)√x²+x dx
= 1/2(2x + 10)√x²+x dx = 1/2∫(2x+1−11)√x²+x dx
= 1/2∫√x²+x(2x+1)dx – 11/2∫√x²+x dx
= 1/2∫√t dt – 11/2 ∫x²+x+1/4−1/4 dx
Put x² + x = t ⇒ (2x + 1) dx = dt

Que-20: ∫(x+3)√3−4x−x² dx

Sol: Let I =∫(x+3)√3−4x−x² dx
= – 1/2∫(−2x−6)√3−4x−x² dx = –1/2∫(−2x−4−2)√3−4x−x² dx
= – 1/2∫√3−4x−x²(−2x−4)dx + ∫√3−4x−x² dx
= – 1/2∫√t dt + ∫−(x²+4x−3) dx
[Put 3 – 4x – x² = t ⇒ (-4 – 2x) dx = dt]

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

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

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