Indefinite Integrals Class 12 OP Malhotra Exe-15E ISC Maths Solutions. In this article you would learn about the integrals of the form ∫dx/√(ax²+bx+c) (1/√(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-15E ISC Maths Solutions Ch-15
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 12th |
| Chapter-15 | Indefinite Integrals |
| Writer | OP Malhotra |
| Exe-15(e) | integrals of the form ∫dx/√(ax²+bx+c) (1/√(quadratic) |
Integrals of the form ∫dx/√(ax²+bx+c) (1/√(quadratic)
Indefinite Integrals Class 12 OP Malhotra Exe-15E Solutions
Que-1: ∫dx/√7−6x−x²
Sol: Let I = ∫dx/√7−6x−x²
= ∫dx/√−(x²+6x−7)
= ∫dx/√−(x²+6x+9−16)
= ∫dx/√16−(x+3)²
Put x + 3 = t ⇒ dx = dt
= ∫dt/√4²−t² = sin-1(t/4) + C
= sin-1(x+3/4) + C
Que-2: ∫dx/√10−8x−2x²
Sol: Let I = ∫dx/√10−8x−2x²
= ∫dx/√−2(x²+4x−5)
= 1/√2∫dx/√−(x²+4x+4−9)
= 1/√2∫dx/√9−(x+2)²
Put x + 2 = t ⇒ dx = dt
= 1/√2∫dt/√3²−t²
= 1/√2 sin-1(x+2/3) + C
Que-3: ∫dx/√4−2x−2x²
Sol:
Que-4: ∫dx/√16−2x−2x²
Sol:
Que-5: ∫ex/√5−4ex−e2x dx
Sol: Let I = ∫ex/√5−4ex−e2x dx
Put ex = t ⇒ ex dx = dt
= ∫dt/√5−4t−t² = ∫dt/√−(t²+4t−5)
= ∫dx/√−(t²+4t+4−9)
= ∫dx/√−{(t+2)²−9}
= ∫dt/√3²−(t+2)²
Put t + 2 = u ⇒ dt = du
= ∫du/√3²−u² = sin-1(u/3) + C
= sin-1(t+2/3) + C
Que-6: ∫1/√(x−1)(x−2) dx
Sol:
Que-7: ∫1/√5x²−2x dx
Sol:
Que-8: ∫2x+1/√x²+2x−1 dx
Sol: Let I = ∫2x+1/√x²+2x−1 dx
= ∫2x+2-1/√x²+2x−1 dx
= ∫(2x+2)/√x²+2x−1 dx – ∫dx/√x²+2x−1
Put x² + 2x – 1 = t in first integral ⇒ (2x + 2)dx = dt
Que-9: ∫x/√8+x−x² dx
Sol: Let I = ∫x/√8+x−x² dx = 1/-2∫(1−2x−1)/√8+x−x² dx
= –1/2∫(1−2x)/√8+x−x² dx + 1/2∫dx/√−(x²−x−8)
Put 8 + x – x² = t in first integral ⇒ (1 – 2x)dx = dt
Que-10: ∫6x+7/√(x−5)(x−4) dx
Sol:
Que-11: ∫4x+1/√2x²+x−3 dx
Sol: Let I = ∫4x+1/√2x²+x−3 dx
Put 2x² + x – 3 = t ⇒ (4x + 1)dx = dt
= ∫dt/√t = ∫t-1/2 dt = t-1/2+1/−1/2+1 + C
= 2√t + C = 2√2x²+x−3 + C
Que-12: ∫x/√x²+x+1 dx
Sol: Let I = ∫x/√x²+x+1 dx
= 1/2∫2x+1−1/√x²+x+1
= 1/2∫2x+1/√x²+x+1 dx – 1/2∫dx/√x²+x+1
Put x² + x + 1 = t in first integral ⇒ (2x + 1)dx = dt
Que-13: ∫x+2/√x²−1 dx
Sol: Let I = ∫x+2/√x²−1 dx
= ∫x/√x²−1 dx + 2∫dx/√x²−1²
Put x² – 1 = t ⇒ 2xdx = dt
Que-14: ∫√5−x/x−2 dx
Sol:
Que-15: ∫5x+3/√x²+4x+10 dx
Sol:
Que-16: ∫x+2/√x²+5x+6 dx
Sol:
–: End of Indefinite Integrals Class 12 OP Malhotra Exe-15E ISC Math Ch-15 Solution :–
Return to :- OP Malhotra ISC Class-12 S Chand Publication Maths Solutions
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