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If f (x) = ∫dx / [(1 + x²)^3/2] and f (0) = 0, then the value of f (1) is

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

If n ∈ N and I_n = ∫(log x)ⁿ dx, then I_n + nI_n-1 =

1) [log x]n+1 / (n + 1) 2) x (log x)n + c 3) (log x)n-1 4) [(log x)n / n] Solution: (2) x (log x)n + c... View Article

∫{e^x [(x² + 1) / (x + 1)²]} dx =

1) [- ex / (x + 1)] + c 2) [ex / (x + 1)] + c 3) [xex / (x + 1)] + c 4) {ex [(x - 1) / (x + 1)]} + c Solution: (4) {ex [(x - 1) / (x + 1)]} + c... View Article

∫{(2 dx) / [(e^x + e^-x)²]} =

1) [- e-x / (ex + e-x)] + c 2) [1 / (ex + e-x)] + c 3) [1 / (ex + 1)2] + c 4) [1 / (ex + e-x)2] + c Solution: (1) [- e-x / (ex + e-x)] + c... View Article

If ∫[(5 tan x) / (tan x – 2)] dx = x + a ln |sin x – 2 cos x| + k, then a =

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

∫{sec² x / [(sec x + tan x)^9/2]} dx = (for some arbitrary constant c)

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

∫{(x² – 1) / [x³ √2x⁴ – 2x² + 1]} dx =

1) (- 1 / 2) √[2 - (2 / x2) + (1 / x4)] + c 2) (1 / 2) log (2x4 - 2x2 + 1) + c 3) (1 / 2) √[2 - (2 / x2) + (1 / x4)] + c 4) None of these... View Article

If ∫f (x) dx = Ѱ (x), then ∫x⁵ f (x³) dx =

1) (1 / 3) [x3 Ѱ (x3) - ∫x2 Ѱ (x3) dx] + c 2) (1 / 3) x3 Ѱ (x3) - 3 ∫x2 Ѱ (x3) dx + c 3) (1 / 3) x3 Ѱ (x3) - ∫x2 Ѱ (x3) dx + c 4) (1 / 3) [x3... View Article

∫{1 / [x³ (x³ – 1)]} dx =

1) (1 / 3) log |(x3) / (x3 - 1)| + c 2) (1 / 3) log |(1 - x3) / (x3)| + c 3) log |(x3) / (x3 - 1)| + c 4) (1 / 3) log |x - 1| + (1 / 2x2) - (1... View Article

∫sec³ x dx =

∫[sec² (sin^-1 x) / (√1 – x²)] dx =

1) sin (tan-1 x) + c 2) tan (sec-1 x) + c 3) tan (sin-1 x) + c 4) - tan (cos-1 x) + c Solution: (3) tan (sin-1 x) + c ∫[sec2 (sin-1 x) / (√1... View Article

∫[(x dx) / (1 + x)^3/2] =

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

For any integer n ≥ 2, let l_n = ∫tanⁿ x dx. If l_n = (1 / a) tan^n-1 x – bl_n-2 for n ≥ 2, then the ordered pair (a, b) =

1) [(n - 1), (n - 1) / (n - 2)] 2) [(n - 1), (n - 2) / (n - 1)] 3) (n, 1) 4) (n - 1, 1) Solution: (4) (n - 1, 1)... View Article

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Archives

If f (x) = ∫dx / [(1 + x²)^3/2] and f (0) = 0, then the value of f (1) is

If n ∈ N and I_n = ∫(log x)ⁿ dx, then I_n + nI_n-1 =

∫{e^x [(x² + 1) / (x + 1)²]} dx =

∫{(2 dx) / [(e^x + e^-x)²]} =

If ∫[(5 tan x) / (tan x – 2)] dx = x + a ln |sin x – 2 cos x| + k, then a =

∫{sec² x / [(sec x + tan x)^9/2]} dx = (for some arbitrary constant c)

∫{(x² – 1) / [x³ √2x⁴ – 2x² + 1]} dx =

If ∫f (x) dx = Ѱ (x), then ∫x⁵ f (x³) dx =

∫{1 / [x³ (x³ – 1)]} dx =

∫sec³ x dx =

∫[sec² (sin^-1 x) / (√1 – x²)] dx =

∫[(x dx) / (1 + x)^3/2] =

For any integer n ≥ 2, let l_n = ∫tanⁿ x dx. If l_n = (1 / a) tan^n-1 x – bl_n-2 for n ≥ 2, then the ordered pair (a, b) =

If ∫dx / [√(sin³ x cos x)] = g (x) + c, then g (x) =

∫{(√(5 + x²)) / (x⁴)} dx =

∫e^x (sin x + 2 cos x) sin x dx =

∫[xe^x / (1 + x)²] dx =

∫[(1 + x)^{e^x} / sin² (xe^x)] dx =

∫{1 / [x² (x⁴ + 1)^¾]} dx =

If linear function f (x) and g (x) satisfy ∫[(3x – 1) cos x + (1 – 2x) sin x] dx = f (x) cos x + g (x) sin x + c, then

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