NDA Indefinite Integrals previous year questions with solutions

Answer: (a) (10e)^x1 + ln 10 + c

Method. e^{x ln 10} = 10^x, so the integrand is (10e)^x

e^{x ln 10} = 10^x, so e^{x ln 10} e^x = 10^x e^x = (10e)^x.
∫ (10e)^x dx = (10e)^xln (10e) + c, and ln (10e) = ln 10 + 1.
So the integral is (10e)^x1 + ln 10 + c, option (a).

Answer: (d) x² + x⁴2 + x⁶3 + C

Method. integrate with respect to t = x²

Let t = x²; then f = 1 + t + t².
∫ (1 + t + t²) dt = t + t²2 + t³3 + C.
= x² + x⁴2 + x⁶3 + C, option (d).

Answer: (b) tan^-1 x + c

Method. sec²(tan^-1 x) = 1 + tan²(tan^-1 x) = 1 + x²

sec² θ = 1 + tan² θ with θ = tan^-1 x gives sec²(tan^-1 x) = 1 + x².
So ∫ dxsec²(tan^-1 x) = ∫ dx1 + x² = tan^-1 x + c.
Hence option (b).

Answer: (d) x⁵5 + c

Method. laws of logarithms: e^{ln f} = f

2 ln x + ln x² = ln x² + ln x² = ln x⁴.
So e^{(2 ln x + ln x2)} = e^{ln x4} = x⁴.
∫ x⁴ dx = x⁵5 + c, option (d).

Answer: (b) p(x)2 (1 + 2 ln 2) + c

Method. ∫ a^kx dx = a^kx/⁠(k ln a)

p(x) = (4e)^2x, so ∫ p(x) dx = (4e)^2x2 ln (4e) + c.
ln (4e) = ln 4 + ln e = 2 ln 2 + 1.
So ∫ p(x) dx = p(x)2(1 + 2 ln 2) + c, option (b).

Answer: (c) 2

Method. the log term must vanish for the integral to be rational

3x² + 8 − 4kx = 3x + 8 − 4kx.
∫ (3x + 8 − 4kx) dx = 3x²2 + (8 − 4k) ln |x| + c.
ln |x| is not a rational function, so its coefficient must be zero: 8 − 4k = 0, k = 2.
Hence option (c).

Answer: (a) a^xln(a) + c

Method. e^{x ln a} = a^x

e^{x ln(a)} = e^ln(ax) = a^x.
∫ a^x dx = a^xln(a) + c, option (a).

Answer: (b) ln |sec x| + c

Method. e^{ln f} = f, then the standard integral of tan x

e^{ln(tan x)} = tan x (where tan x > 0).
∫ tan x dx = ∫ sin xcos x dx = -ln |cos x| + c = ln |sec x| + c.
Hence option (b).

Answer: (b) ln (1 − 2^-x)ln 2 + c

Method. divide numerator and denominator by 2^x, then substitute u = 1 − 2^-x

12^x - 1 = 2^-x1 − 2^-x.
Put u = 1 − 2^-x: du = 2^-x ln 2 dx, so 2^-x dx = duln 2.
∫ 2^-x dx1 − 2^-x = 1ln 2 ∫ duu = ln (1 − 2^-x)ln 2 + c (taking x > 0).
Hence option (b).

Answer: (d) Statement 1 is false but Statement 2 is true

Method. integrate sin² x and test the period of the antiderivative

sin² x = 1 − cos 2x2, so f(x) = x2 − sin 2x4 + c.
f(x + π) = x + π2 − sin (2x + 2π)4 + c = f(x) + π2 ≠ f(x). Statement 1 is false.
sin (x + π) = -sin x, so sin²(x + π) = sin² x. Statement 2 is true.
Hence option (d).

Shortcut. A periodic integrand with non-zero average (sin² x averages 1/⁠2) has a non-periodic antiderivative.

Answer: (d) 23 sin^-1(x^3/⁠2) + c

Method. substitution t = x^3/⁠2 gives ∫ dt/⁠√(1⁠ −⁠ t²)

Put t = x^3/⁠2; then t² = x³ and dt = 32 ⁠√x⁠ dx, so ⁠√x⁠ dx = 23 dt.
∫ ⁠√x⁠ dx⁠√(1⁠ −⁠ x³) = 23 ∫ dt⁠√(1⁠ −⁠ t²) = 23 sin^-1 t + c = 23 sin^-1(x^3/⁠2) + c.
Hence option (d).

Answer: (c) 1ln 10 tan^-1(10^x) + c

Method. multiply by 10^x/⁠10^x and substitute t = 10^x

110^x + 10^-x = 10^x10^2x + 1.
Put t = 10^x, dt = 10^x ln 10 dx: ∫ 10^x dx10^2x + 1 = 1ln 10 ∫ dtt² + 1 = 1ln 10 tan^-1 t + c.
= 1ln 10 tan^-1(10^x) + c, option (c).

Answer: (b) 12 x^2x + c

Method. the integrand is half the derivative of x^2x

Let y = x^2x; ln y = 2x ln x, so 1y dydx = 2(1 + ln x), i.e. ddx x^2x = 2x^2x(1 + ln x).
(x^x)² = x^2x, so the integrand is x^2x(1 + ln x) = 12 ddx x^2x.
Hence ∫ (x^x)²(1 + ln x) dx = 12 x^2x + c, option (b).

Answer: (b) x + c

Method. bring both integrands to the denominator e^2x + 1 and add

Multiply numerator and denominator of I₁'s integrand by e^x: e^xe^x + e^-x = e^2xe^2x + 1.
So I₁ + I₂ = ∫ e^2x + 1e^2x + 1 dx = ∫ dx = x + c.
Hence option (b).

Answer: (a) 12 ln(x²x² + 1) + C

Method. multiply by x/⁠x and substitute t = x²

∫ dxx(x² + 1) = ∫ x dxx²(x² + 1); put t = x², dt = 2x dx.
= 12 ∫ dtt(t + 1) = 12 ∫ (1t − 1t + 1) dt = 12 ln (tt + 1) + C.
= 12 ln(x²x² + 1) + C, option (a).

Answer: (a) 1n ln (xⁿxⁿ + 1) + c

Method. multiply by x^n-1/⁠x^n-1 and substitute t = xⁿ

∫ dxx(xⁿ + 1) = ∫ x^n-1 dxxⁿ(xⁿ + 1).
Put t = xⁿ, dt = n x^n-1 dx: 1n ∫ dtt(t + 1) = 1n ∫ (1t − 1t + 1) dt.
= 1n ln (tt + 1) + c = 1n ln (xⁿxⁿ + 1) + c.
Hence option (a).

Answer: (d) -1(n − 1)(1 + ln x)^n-1 + c

Method. substitution t = 1 + ln x, then the power rule

Put t = 1 + ln x, dt = dxx.
∫ dxx(1 + ln x)ⁿ = ∫ t^-n dt = t^1-n1 − n + c (n ≠ 1).
t^1-n1 − n = -1(n − 1) t^n-1.
So the integral is -1(n − 1)(1 + ln x)^n-1 + c, option (d).

Answer: (d) 17 ln |x⁷x⁷ + 1| + c

Method. multiply by x⁶/⁠x⁶ and substitute t = x⁷

∫ dxx(x⁷ + 1) = ∫ x⁶ dxx⁷(x⁷ + 1).
Put t = x⁷, dt = 7x⁶ dx: the integral becomes 17 ∫ dtt(t + 1).
1t(t + 1) = 1t − 1t + 1, so 17 ∫ dtt(t + 1) = 17 ln |tt + 1| + c.
Hence the answer is 17 ln |x⁷x⁷ + 1| + c, option (d).

Shortcut. For ∫ dxx(xⁿ + 1) the answer is always 1n ln |xⁿxⁿ + 1| + c.

Answer: (d) 1e ln (x^e + e^x) + c

Method. numerator is (1/⁠e) times the derivative of the denominator

ddx(x^e + e^x) = e x^e-1 + e^x = e(x^e-1 + e^x-1).
So the numerator x^e-1 + e^x-1 = 1e ddx(x^e + e^x).
Using ∫ f'(x)f(x) dx = ln |f(x)| + c, the integral is 1e ln (x^e + e^x) + c, option (d).

Answer: (d) 26

Method. numerator = λ(denominator) + μ(derivative of denominator)

Write the numerator as λ(denominator) + μ(derivative of denominator):
3 cos x + 4 sin x = λ(2 cos x + 5 sin x) + μ(5 cos x − 2 sin x).
cos x: 2λ + 5μ = 3; sin x: 5λ − 2μ = 4. Solving, λ = 2629, μ = ⁠729⁠.
So the integral is λx + μ ln |2 cos x + 5 sin x| + c = 26x29 + 729 ln |2 cos x + 5 sin x| + c; α = 26, β = 7.
Hence α = 26, option (d).

Answer: (a) 7

Method. numerator = λ(denominator) + μ(derivative of denominator)

Write the numerator as λ(denominator) + μ(derivative of denominator):
3 cos x + 4 sin x = λ(2 cos x + 5 sin x) + μ(5 cos x − 2 sin x).
cos x: 2λ + 5μ = 3; sin x: 5λ − 2μ = 4. Solving, λ = 2629, μ = ⁠729⁠.
So the integral is λx + μ ln |2 cos x + 5 sin x| + c = 26x29 + 729 ln |2 cos x + 5 sin x| + c; α = 26, β = 7.
Hence β = 7, option (a).

Answer: (c) 2⁠√(sin⁠ x) + 2⁠√(cos⁠ x) + c

Method. split the fraction into cos x/⁠√(sin⁠ x) − sin x/⁠√(cos⁠ x)

(cos x)^3/⁠2⁠√(sin⁠ x cos x) = cos x⁠√(sin⁠ x) and (sin x)^3/⁠2⁠√(sin⁠ x cos x) = sin x⁠√(cos⁠ x).
∫ cos x⁠√(sin⁠ x) dx = 2⁠√(sin⁠ x) (put t = sin x).
∫ sin x⁠√(cos⁠ x) dx = -2⁠√(cos⁠ x) (put t = cos x).
So the integral is 2⁠√(sin⁠ x) − (-2⁠√(cos⁠ x)) + c = 2⁠√(sin⁠ x) + 2⁠√(cos⁠ x) + c, option (c).

Answer: (b) 2⁠√(tan⁠ x) + c

Method. write the integrand as (tan x)^-1/⁠2 sec² x and substitute t = tan x

(sin x)^-1/⁠2(cos x)^-3/⁠2 = (sin xcos x)^-1/⁠2 · 1cos² x = (tan x)^-1/⁠2 sec² x.
Put t = tan x, dt = sec² x dx: ∫ t^-1/⁠2 dt = 2t^1/⁠2 + c = 2⁠√(tan⁠ x) + c.
Hence option (b).

Answer: (b) A + B − 2 = 0

Method. 1 − sin 2x = (cos x − sin x)², with cos x > sin x on (0, π/⁠4)

1 − sin 2x = sin² x + cos² x − 2 sin x cos x = (cos x − sin x)².
For 0 < x < π4, cos x > sin x, so ⁠√(1⁠ −⁠ sin 2x) = cos x − sin x.
∫ (cos x − sin x) dx = sin x + cos x + C, so A = 1, B = 1.
A + B = 2, i.e. A + B − 2 = 0, option (b).

Answer: (d) ln |sec x + tan x| − ln |sec x| + c

Method. rewrite in sin and cos: the integrand is f'/⁠f with f = 1 + sin x

1sec x + tan x = cos x1 + sin x.
∫ cos x dx1 + sin x = ln |1 + sin x| + c.
In option (d): ln |sec x + tan x| − ln |sec x| = ln |sec x + tan xsec x| = ln |1 + sin x|.
Hence option (d).

Answer: (d) (1 − cos² x)²4 + c

Method. substitution t = sin x

Put t = sin x, dt = cos x dx: ∫ sin³ x cos x dx = ∫ t³ dt = t⁴4 + c = sin⁴ x4 + c.
Since sin² x = 1 − cos² x, sin⁴ x4 = (1 − cos² x)²4.
Hence option (d).

Answer: (a) c + 1ab tan^-1(a tan xb)

Method. divide by cos² x and substitute t = tan x

Divide numerator and denominator by cos² x: ∫ sec² x dxa² tan² x + b².
Put t = tan x, dt = sec² x dx: ∫ dta² t² + b² = 1a² ∫ dtt² + (b/⁠a)² = 1a² · ab tan^-1(atb) + c.
= c + 1ab tan^-1(a tan xb), option (a).

Answer: (a) πx4 + x²4 + c

Method. write sec x + tan x as tan(π/⁠4 + x/⁠2)

sec x + tan x = 1 + sin xcos x = (cos x2 + sin x2)²cos² x2 − sin² x2 = cos x2 + sin x2cos x2 − sin x2.
Dividing by cos x2: 1 + tan x21 − tan x2 = tan(π4 + x2).
For -π2 < x < π2, tan^-1(sec x + tan x) = π4 + x2.
∫ (π4 + x2) dx = πx4 + x²4 + c, option (a).

Answer: (d) dvdx

Method. evaluate u and v by parts, then compare with their derivatives

By parts twice (or by the standard result ∫ e^ax cos bx dx), u = e^x(cos x + sin x)2 and v = e^x(sin x − cos x)2, taking the constants as zero.
By definition, dudx = e^x cos x and dvdx = e^x sin x.
u + v = e^x(2 sin x)2 = e^x sin x = dvdx, option (d).
With arbitrary constants u + v = e^x sin x + C; (d) is the only option that differs from u + v by a constant, so the answer does not depend on the constants chosen.

Answer: (d) Neither I nor II

Method. evaluate u and v by parts, then compare with their derivatives

By parts twice (or by the standard result ∫ e^ax cos bx dx), u = e^x(cos x + sin x)2 and v = e^x(sin x − cos x)2, taking the constants as zero.
By definition, dudx = e^x cos x and dvdx = e^x sin x.
I: dudx = e^x cos x, but -v = e^x(cos x − sin x)2; not equal, so I is false.
II: dvdx = e^x sin x, but -u = -e^x(cos x + sin x)2; not equal, so II is false.
Hence neither, option (d).

Answer: (a) xe^x ln x + c

Method. ∫ e^x{f(x) + f'(x)} dx = e^x f(x) + c with f(x) = x ln x

Take f(x) = x ln x; then f'(x) = ln x + 1.
f(x) + f'(x) = x ln x + ln x + 1, which is exactly the bracket.
So ∫ e^x{1 + ln x + x ln x} dx = e^x · x ln x + c = xe^x ln x + c, option (a).

Answer: (c) x sin x + cos x + sin² x2 + c

Method. e^{log x} = x, then integration by parts for x cos x

e^{log x} = x, so the integrand is x cos x + sin x cos x.
∫ x cos x dx = x sin x - ∫ sin x dx = x sin x + cos x.
∫ sin x cos x dx = sin² x2 (put t = sin x).
Total: x sin x + cos x + sin² x2 + c, option (c).

Answer: (a) 2x ln(x) − 2x + c

Method. ln(x²) = 2 ln x, then ∫ ln x dx by parts

For x > 0, ln(x²) = 2 ln x.
∫ ln x dx = x ln x - ∫ x · 1x dx = x ln x − x.
So ∫ ln(x²) dx = 2x ln(x) − 2x + c, option (a).

Answer: (a) x(ln x)^-1 + c

Method. integrate (ln x)^-1 by parts; the (ln x)^-2 terms cancel

Take 1 as the second function: ∫ (ln x)^-1 dx = x(ln x)^-1 - ∫ x · (-(ln x)^-2 · 1x) dx = x(ln x)^-1 + ∫ (ln x)^-2 dx.
So ∫ (ln x)^-1 dx - ∫ (ln x)^-2 dx = x(ln x)^-1 + c, option (a).

Shortcut. Differentiate the options: ddx xln x = 1ln x − 1(ln x)².

Answer: (c) 15

Method. substitution t = cos θ, then partial fractions

Put t = cos θ, dt = -sin θ dθ: the integral becomes -∫ dt(2 + t)(3 + 4t).
Partial fractions: 1(2 + t)(3 + 4t) = P2 + t + Q3 + 4t; t = -2 gives P = 13 − 8 = -15; t = -34 gives Q = 12 − 3/⁠4 = 45.
So -∫ [-15 · 12 + t + 45 · 13 + 4t] dt = 15 ln |2 + t| − 45 · 14 ln |3 + 4t|.
= 15 ln |2 + cos θ| − 15 ln |3 + 4 cos θ| + c, so A = 15, B = -15.
Hence A = 15, option (c).

Answer: (b) -15

Method. substitution t = cos θ, then partial fractions

Put t = cos θ, dt = -sin θ dθ: the integral becomes -∫ dt(2 + t)(3 + 4t).
Partial fractions: 1(2 + t)(3 + 4t) = P2 + t + Q3 + 4t; t = -2 gives P = 13 − 8 = -15; t = -34 gives Q = 12 − 3/⁠4 = 45.
So -∫ [-15 · 12 + t + 45 · 13 + 4t] dt = 15 ln |2 + t| − 45 · 14 ln |3 + 4t|.
= 15 ln |2 + cos θ| − 15 ln |3 + 4 cos θ| + c, so A = 15, B = -15.
Hence B = -15, option (b).

Answer: (b) 1

Method. split x² - 1 = (x² + 1) − 2 and use the standard ⁠√(x²⁠ + a²) integrals

Write x² - 1 = (x² + 1) − 2:
2∫ x² - 1⁠√(x²⁠ + 1) dx = 2∫ ⁠√(x²⁠ + 1) dx − 4∫ dx⁠√(x²⁠ + 1).
∫ ⁠√(x²⁠ + 1) dx = x2⁠√(x²⁠ + 1) + 12 ln (x + ⁠√(x²⁠ + 1)); ∫ dx⁠√(x²⁠ + 1) = ln (x + ⁠√(x²⁠ + 1)).
So the left side is x⁠√(x²⁠ + 1) + ln (x + ⁠√(x²⁠ + 1)) − 4 ln (x + ⁠√(x²⁠ + 1)) + c = x⁠√(x²⁠ + 1) − 3 ln (x + ⁠√(x²⁠ + 1)) + c.
Comparing with U(x)V(x) − 3 ln {U(x) + V(x)}: U(x) = x, V(x) = ⁠√(x²⁠ + 1).
|U² - V²| = |x² - (x² + 1)| = 1, option (b).

Answer: (a) ⁠√(x²⁠ + x⁴)

Method. split x² - 1 = (x² + 1) − 2 and use the standard ⁠√(x²⁠ + a²) integrals

Write x² - 1 = (x² + 1) − 2:
2∫ x² - 1⁠√(x²⁠ + 1) dx = 2∫ ⁠√(x²⁠ + 1) dx − 4∫ dx⁠√(x²⁠ + 1).
∫ ⁠√(x²⁠ + 1) dx = x2⁠√(x²⁠ + 1) + 12 ln (x + ⁠√(x²⁠ + 1)); ∫ dx⁠√(x²⁠ + 1) = ln (x + ⁠√(x²⁠ + 1)).
So the left side is x⁠√(x²⁠ + 1) + ln (x + ⁠√(x²⁠ + 1)) − 4 ln (x + ⁠√(x²⁠ + 1)) + c = x⁠√(x²⁠ + 1) − 3 ln (x + ⁠√(x²⁠ + 1)) + c.
Comparing with U(x)V(x) − 3 ln {U(x) + V(x)}: U(x) = x, V(x) = ⁠√(x²⁠ + 1).
U(x)V(x) = x⁠√(x²⁠ + 1). Since ⁠√(x²⁠ + x⁴) = |x|⁠√(x²⁠ + 1), this equals option (a) for x ≥ 0 (the options presume x ≥ 0: no option equals x⁠√(x²⁠ + 1) for x < 0, and (b), (c), (d) do not equal it for x > 0 either). Hence option (a).

Answer: (a) 13

Method. rationalise the denominator

Rationalise: 1⁠√(x⁠ + 1) - ⁠√(x⁠ −⁠ 1) = ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)(x + 1) − (x − 1) = ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)2.
∫ ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)2 dx = 12 · 23[(x + 1)^3/⁠2 + (x − 1)^3/⁠2] + c = 13(x + 1)^3/⁠2 + 13(x − 1)^3/⁠2 + c.
So α = 13, β = 13.
Hence α = 13, option (a).

Answer: (c) 13

Method. rationalise the denominator

Rationalise: 1⁠√(x⁠ + 1) - ⁠√(x⁠ −⁠ 1) = ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)(x + 1) − (x − 1) = ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)2.
∫ ⁠√(x⁠ + 1) + ⁠√(x⁠ −⁠ 1)2 dx = 12 · 23[(x + 1)^3/⁠2 + (x − 1)^3/⁠2] + c = 13(x + 1)^3/⁠2 + 13(x − 1)^3/⁠2 + c.
So α = 13, β = 13.
Hence β = 13, option (c).

Answer: (d) tan^-1(2x − 1) + c

Method. complete the square to reach ∫ dt/⁠(1 + t²)

2x² - 2x + 1 = (2x − 1)² + 12.
So ∫ dx2x² - 2x + 1 = ∫ 2 dx(2x − 1)² + 1.
Put t = 2x − 1, dt = 2 dx: ∫ dtt² + 1 = tan^-1 t + c = tan^-1(2x − 1) + c.
Hence option (d).