JEE Mathematics Indefinite Integration MCQs Set 23

Multiple Choice Questions (MCQs) for JEE Mathematics: Indefinite Integration

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Multiple Choice Questions

Question 1: The integral of \( f(x)=\sqrt{1+x^2} \) with respect to x² is
(a) \( \frac23\frac{(1+x^2)^{3/2}}{x}+k \)
(b) \( \frac23(1+x^2)^{3/2}+k \)
(c) \( \frac23x(1+x^2)^{3/2}+k \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac23(1+x^2)^{3/2}+k \)

Explanation:
1. Put t = x²: ∫√(1 + t) dt = (2/3)(1 + t)^(3/2) + k.
2. That is (2/3)(1 + x²)^(3/2) + k.

Teacher's Note:
1. Integrating with respect to x² means using t = x² as the variable.

Question 2: \( \int\frac{d(x^2+1)}{\sqrt{x^2+2}} \) is equal to
(a) \( 2\sqrt{x^2+2}+k \)
(b) \( \sqrt{x^2+2}+k \)
(c) \( \frac{1}{(x^2+2)^{3/2}}+k \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( 2\sqrt{x^2+2}+k \)

Explanation:
1. d(x² + 1) = 2x dx.
2. ∫2x dx/√(x² + 2) = 2√(x² + 2) + k.

Teacher's Note:
1. Differentiate the bracket first.

Question 3: \( \int x^x(1+\log x)\,dx \) is equal to
(a) xˣ log x + k
(b) \( e^{x^x}+k \)
(c) xˣ + k
(d) none of these
Show Answer & Explanation

Answer: (c) xˣ + k

Explanation:
1. d/dx (xˣ) = xˣ(1 + ln x).
2. So the integral is xˣ + k.

Teacher's Note:
1. Logarithmic differentiation of xˣ.

Question 4: \( \int\frac{x\,dx}{1+x^4} \) is equal to
(a) \( \tan^{-1}x^2+k \)
(b) \( \frac12\tan^{-1}x^2+k \)
(c) \( \log(1+x^4)+k \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac12\tan^{-1}x^2+k \)

Explanation:
1. Put t = x², dt = 2x dx.
2. The integral becomes ½∫dt/(1 + t²) = ½ tan⁻¹x² + k.

Teacher's Note:
1. Recognise the integrand as a derivative of a simple expression.
2. Differentiate your answer to verify it.

Question 5: The antiderivative of \( \frac{2^x}{\sqrt{1-4^x}} \) with respect to x is
(a) \( \log_2e\cdot\sin^{-1}(2^x)+k \)
(b) \( \sin^{-1}(2^x)+k \)
(c) \( \cos^{-1}(2^x)\cdot\frac{1}{\log_e2}+k \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \log_2e\cdot\sin^{-1}(2^x)+k \)

Explanation:
1. Put t = 2ˣ, dt = 2ˣ ln 2 dx.
2. The integral = (1/ln 2)∫dt/√(1 − t²) = (1/ln 2) sin⁻¹(2ˣ) = log₂e · sin⁻¹(2ˣ).

Teacher's Note:
1. Note 4ˣ = (2ˣ)².

Question 6: \( \int\frac{(1+x)^2}{x+x^3}dx \) is equal to
(a) logₑx + logₑ(1 + x²) + k
(b) logₑx + tan⁻¹x + k
(c) logₑx + 2tan⁻¹x + k
(d) none of these
Show Answer & Explanation

Answer: (c) logₑx + 2tan⁻¹x + k

Explanation:
1. (1 + x)²/(x(1 + x²)) = (1 + x² + 2x)/(x(1 + x²)) = 1/x + 2/(1 + x²).
2. The integral is ln x + 2tan⁻¹x + k.

Teacher's Note:
1. Split the numerator as (1 + x²) + 2x.

Question 7: \( \int\frac{dx}{x^{1/5}(1+x^{4/5})^{1/2}} \) is
(a) \( \sqrt{1+x^{4/5}}+k \)
(b) \( \frac52\sqrt{1+x^{4/5}}+k \)
(c) \( x^{4/5}(1+x^{4/5})^{1/2}+k \)
(d) none of these
Show Answer & Explanation

Answer: (b) \( \frac52\sqrt{1+x^{4/5}}+k \)

Explanation:
1. Put u = x^(4/5), du = (4/5)x^(−1/5)dx.
2. The integral = (5/4)∫du/√(1 + u) = (5/4)·2√(1 + u) = (5/2)√(1 + x^(4/5)) + k.

Teacher's Note:
1. Recognise the integrand as a derivative of a simple expression.
2. Differentiate your answer to verify it.

Question 8: The primitive of the function x|cos x| when π/2 < x < π is given by
(a) cos x + x sin x
(b) −cos x − x sin x
(c) x sin x − cos x
(d) none of these
Show Answer & Explanation

Answer: (b) −cos x − x sin x

Explanation:
1. For π/2 < x < π, cos x < 0, so |cos x| = −cos x.
2. ∫−x cos x dx = −(x sin x + cos x) = −cos x − x sin x.

Teacher's Note:
1. Remove the modulus using the sign of cos x on the interval.

Question 9: \( \int x\sec x^2\,dx \) is equal to
(a) \( \frac12\log(\sec x^2+\tan x^2)+k \)
(b) \( \frac{x^2}{2}\log(\sec x^2+\tan x^2)+k \)
(c) \( 2\log(\sec x^2+\tan x^2)+k \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \frac12\log(\sec x^2+\tan x^2)+k \)

Explanation:
1. Put t = x²: ½∫sec t dt = ½ ln(sec t + tan t) + k.

Teacher's Note:
1. Recognise the integrand as a derivative of a simple expression.
2. Differentiate your answer to verify it.

Question 10: \( \int\sin2x\cdot\log\cos x\,dx \) is equal to
(a) \( \cos^2x\left(\frac12+\log\cos x\right)+k \)
(b) \( \cos^2x\cdot\log\cos x+k \)
(c) \( \cos^2x\left(\frac12-\log\cos x\right)+k \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \cos^2x\left(\frac12-\log\cos x\right)+k \)

Explanation:
1. Put t = cos²x, so sin 2x dx = −dt and log cos x = ½ ln t.
2. The integral becomes −½∫ln t dt = −½(t ln t − t) + k = (t/2)(1 − ln t) + k.
3. Since ln t = 2 ln cos x, this is (cos²x/2)(1 − 2 ln cos x) = cos²x(1/2 − ln cos x) + k.

Teacher's Note:
1. Use the substitution t = cos²x and integrate ln t by parts.

Question 11: \( \int\frac{1+\sin x}{1+\cos x}e^x\,dx \) is equal to
(a) \( e^x\tan\frac x2+k \)
(b) \( e^x\tan x+k \)
(c) \( \frac12e^x\tan\frac x2+k \)
(d) \( e^x\sec^2\frac x2+k \)
Show Answer & Explanation

Answer: (a) \( e^x\tan\frac x2+k \)

Explanation:
1. (1 + sin x)/(1 + cos x) = ½sec²(x/2) + tan(x/2).
2. With f = tan(x/2), f′ = ½sec²(x/2), the integral is ∫eˣ(f + f′)dx = eˣ tan(x/2) + k.

Teacher's Note:
1. Use ∫eˣ(f + f′)dx = eˣ f.

Question 12: If f(0) = f′(0) = 0 and f″(x) = tan²x, then f(x) is
(a) \( \log\sec x-\frac12x^2 \)
(b) \( \log\cos x+\frac12x^2 \)
(c) \( \log\sec x+\frac12x^2 \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( \log\sec x-\frac12x^2 \)

Explanation:
1. f′(x) = ∫tan²x dx = tan x − x + C, and f′(0) = 0 gives C = 0.
2. f(x) = ln sec x − x²/2 + C₁, and f(0) = 0 gives C₁ = 0.

Teacher's Note:
1. Integrate twice and use both initial conditions.

Question 13: \( \int\frac{dx}{\cos x+\sqrt3\sin x} \) is equal to
(a) \( \log\tan\left(\frac x2+\frac\pi3\right)+k \)
(b) \( \log\tan\left(\frac x2-\frac\pi3\right)+k \)
(c) \( \frac12\log\tan\left(\frac x2+\frac{\pi}{12}\right)+k \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( \frac12\log\tan\left(\frac x2+\frac{\pi}{12}\right)+k \)

Explanation:
1. cos x + √3 sin x = 2cos(x − π/3).
2. ∫dx/(2cos(x − π/3)) = ½ ln tan(π/4 + (x − π/3)/2) = ½ ln tan(x/2 + π/12).

Teacher's Note:
1. Write the denominator as R cos(x − α) and use ∫sec u du.

Question 14: If \( \int\tan^4x\,dx=a\tan^3x+b\tan x+\phi(x) \), consider the statements:
1. a = 1/3
2. b = 1
3. φ(x) = x + c
4. b = −1
(a) 1 and 3 only
(b) All of 1, 2, 3 and 4
(c) 1, 3 and 4 only
(d) 1 and 4 only
Show Answer & Explanation

Answer: (c) 1, 3 and 4 only

Explanation:
1. tan⁴x = tan²x(sec²x − 1) = tan²x sec²x − sec²x + 1.
2. The integral is tan³x/3 − tan x + x + c.
3. So a = 1/3, b = −1 and φ(x) = x + c.

Teacher's Note:
1. Reduce tan⁴x using tan² = sec² − 1 twice.

Question 15: If \( \int x\log(1+x^2)dx=\phi(x)\cdot\log(1+x^2)+\psi(x)+c \), consider the statements:
1. φ(x) = (1 + x²)/2
2. ψ(x) = (1 + x²)/2
3. ψ(x) = −(1 + x²)/2
4. φ(x) = −(1 + x²)/2
(a) 1 and 3 only
(b) 1, 3 and 4 only
(c) 1, 2 and 3 only
(d) All of 1, 2, 3 and 4
Show Answer & Explanation

Answer: (a) 1 and 3 only

Explanation:
1. Put t = 1 + x²: ½∫ln t dt = ½(t ln t − t).
2. So φ = (1 + x²)/2 and ψ = −(1 + x²)/2.

Teacher's Note:
1. Substitute t = 1 + x² and integrate ln t.

Question 16: If \( \int\frac{dx}{(x+1)(x-2)}=A\log(x+1)+B\log(x-2)+C \), consider the statements:
1. A + B = 0
2. AB = −1
3. A : B = −1
(a) All of 1, 2 and 3
(b) 1 and 3 only
(c) Only 1
(d) Only 3
Show Answer & Explanation

Answer: (b) 1 and 3 only

Explanation:
1. 1/((x + 1)(x − 2)) = (1/3)[1/(x − 2) − 1/(x + 1)].
2. So A = −1/3 and B = 1/3: A + B = 0 and A : B = −1, while AB = −1/9.

Teacher's Note:
1. Partial fractions.

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Indefinite Integration Objective Questions & Solutions for JEE Mathematics

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