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If (d / dx) [a tan^-1 x + b log [(x – 1) / (x + 1)] = 1 / (x⁴ – 1), then a – 2b =

1) 1 2) - 1 3) 0 4) 2 Solution: (2) - 1... View Article

If (d / dx) (f (x)) = 1 / (1 + x²), then (d / dx) (f (x³)) =

1) 3x / (1 + x3) 2) 3x2 / (1 + x6) 3) (- 6x5) / (1 + x6)2 4) (- 6x5) / (1 + x6) Solution: (2) 3x2 / (1 + x6) (d / dx) (f (x)) = 1 / (1 + x2)... View Article

∫(x + 1) (x + 2)⁷ (x + 3) dx =

1) [(x + 2)10 / 10] - [(x + 2)8 / 8] + c 2) [(x + 1)2 / 2] - [(x + 2)8 / 8] - [(x + 3)2 / 2] + c 3) [(x + 1)10 / 10] + c 4) [(x + 1)2 / 2] +... View Article

∫cosec (x – a) cosec x dx =

1) (1 / sin a) log |sin (x - a) / sin x| + C 2) (- 1 / sin a) log [sin (x - a) sin x] + C 3) (1 / sin a) log [sin (x - a) cosec x] + C 4) (1 /... View Article

∫(x⁴ + x² + 1) / (x² – x + 1) dx =

1) (x3 / 3) - (x2 / 2) + x + C 2) (x3 / 3) + (x2 / 2) + x + C 3) (x3 / 3) - (x2 / 2) - x + C 4) (x3 / 3) + (x2 / 2) - x + C Solution: (2) (x3... View Article

∫[sin (log x) + cos (log x)] dx =

1) x cos (log x) + C 2) cos (log x) + C 3) x sin (log x) + C 4) sin (log x) + C Solution: (3) x sin (log x) + C ∫[sin (log x) + cos (log x)]... View Article

∫[d (cos θ) / (√1 – cos² θ)] =

1) cos-1 θ + C 2) θ + C 3) sin-1 θ + C 4) sin-1 (cos θ) + C Solution: (4) sin-1 (cos θ) + C I = ∫[d (cos θ) / (√1 - cos2 θ)] = - ∫[sin θ dθ]... View Article

If f [(3x – 4) / (3x + 4)] = x + 2, then ∫f (x) dx =

1) ex+2 log |(3x - 4) / (3x + 4)| + c 2) (- 8 / 3) log |1 - x| + (2 / 3) x + c 3) (8 / 3) log |1 - x| + (x / 3) + c 4) e[(3x - 4) / (3x + 4)]... View Article

∫1 / [sin (x – a) sin (x – b)] dx =

1) 1 / [sin (b - a)] log |[(sin (x + b)) / (sin (x + a))| + c 2) 1 / [sin (b + a)] log |[(sin (x - b)) / (sin (x - a))| - c 3) 1 / [sin (b -... View Article

The value of ∫(sin x + cos x) / [√(1 + sin 2x)] dx is

1) √(1 + cos 2x) + c 2) √x - c 3) x + c 4) √(x + 2x) + c Solution: (3) x + c... View Article

∫(sin⁸ x – cos⁸ x) / (1 – 2 sin² x cos² x) dx =

1) (- 1 / 2) sin 2x + c 2) (1 / 2) sin 2x + c 3) (1 / 2) sin x + c 4) (- 1 / 2) sin x + c Solution: (1) (- 1 / 2) sin 2x + c... View Article

∫sin^-1 (cos 3x) dx =

1) (3π / 2) x - (x2 / 2) + c 2) (π / 2) x - (3x2 / 2) + c 3) (3π / 2) x + (x2 / 2) + c 4) (π / 2) x - (x3 / 2) + c Solution: (2) (π / 2) x -... View Article

If f (x) and g (x) are twice differentiable function on (0, 3) satisfying f ” (x) = g” (x), f ‘ (1) = 4, g’ (1) = 6, f (2) = 3, g (2) = 9, then f (1) – g (1) is equal to

1) 4 2) - 4 3) 0 4) - 2 Solution: (2) - 4 f '' (x) = g'' (x) ∫f’’ (x) dx = ∫g’’ (x) dx At x = 1, f’ = g’ + c 4 = 6 + c c = - 2 f’ = g’ - 1 f... View Article

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Archives

If (d / dx) [a tan^-1 x + b log [(x – 1) / (x + 1)] = 1 / (x⁴ – 1), then a – 2b =

If (d / dx) (f (x)) = 1 / (1 + x²), then (d / dx) (f (x³)) =

∫(x + 1) (x + 2)⁷ (x + 3) dx =

∫cosec (x – a) cosec x dx =

∫(x⁴ + x² + 1) / (x² – x + 1) dx =

∫[sin (log x) + cos (log x)] dx =

∫[d (cos θ) / (√1 – cos² θ)] =

If f [(3x – 4) / (3x + 4)] = x + 2, then ∫f (x) dx =

∫1 / [sin (x – a) sin (x – b)] dx =

The value of ∫(sin x + cos x) / [√(1 + sin 2x)] dx is

∫(sin⁸ x – cos⁸ x) / (1 – 2 sin² x cos² x) dx =

∫sin^-1 (cos 3x) dx =

If f (x) and g (x) are twice differentiable function on (0, 3) satisfying f ” (x) = g” (x), f ‘ (1) = 4, g’ (1) = 6, f (2) = 3, g (2) = 9, then f (1) – g (1) is equal to

If I₁ = ∫sin^-1 x dx and I₂ = ∫sin^-1 √(1 – x²) dx, then

∫1 / (√7 – x²) dx =

The value of ∫(cos 2x – 1) / (cos 2x + 1) dx is

∫a^x dx =

If ∫(x² – 1) / [(x + 1)² √[x (x² + x + 1)]] dx = A tan^-1 [(√x² + x + 1) / (x)] + c where c is real constant, then A =

If ∫dx / [(1 + √x) (√x – x²)] = (A √x / (√1 – x)) + B / (√1 – x) + c where c is real constant, then A + B =

∫√(1 + cos x) dx =

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