JEE Mathematics Indefinite Integration MCQs Set 21

Answer: (b) \( x - \log{\left(x \right)} + \frac{\log{\left(x^{2} + 1 \right)}}{2} - \tan^{-1}{\left(x \right)} + c \)

Explanation:
1. (x³ - 1)/(x(x² + 1)) = 1 - (x + 1)/(x(x² + 1)).
2. (x + 1)/(x(x² + 1)) = 1/x + (1 - x)/(x² + 1) by partial fractions.
3. The integrand is 1 - 1/x + x/(x² + 1) - 1/(x² + 1).
4. Integral = x - log x + (1/2) log(x² + 1) - tan⁻¹x.
5. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Split rational functions into partial fractions, or complete the square for quadratics.
2. Compare coefficients carefully to avoid algebra slips.

Answer: (a) \( 2 \left(- \sqrt{x} + x + 1\right) e^{\sqrt{x}} + c \)

Explanation:
1. Put x = t², dx = 2t dt: the integral is 2∫e^t (t² + t) dt.
2. Write ∫e^t(t² + t) dt = e^t(At² + Bt + C). Differentiating gives A = 1, 2A + B = 1 and B + C = 0, so A = 1, B = -1, C = 1.
3. So the integral is 2e^t(t² - t + 1).
4. Integral = 2e^(√x)(x - √x + 1) + c.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (a) \( 2\tan^2x \)

Explanation:
1. sec²x + tan²x = 1 + 2tan²x.
2. Put 2tan²x = t, so dt = 4 tan x sec²x dx.
3. ∫dt/(1 + t) = log(1 + t) = log(1 + 2tan²x).
4. Hence f(x) = 2tan²x.
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (b) \( \sec^{-1}{\left(\cos{\left(x \right)} + \sec{\left(x \right)} \right)} + c \)

Explanation:
1. 1 + cos²x = cos x (sec x + cos x) and √(1 + cos²x + cos⁴x) = cos x √((sec x + cos x)² - 1), so the integrand is sin³x/(cos²x (sec x + cos x)√((sec x + cos x)² - 1)).
2. With t = sec x + cos x, dt = (sec x tan x - sin x) dx = sin x (1 - cos²x)/cos²x dx = sin³x/cos²x dx.
3. The integral is ∫dt/(t√(t² - 1)) = sec⁻¹t.
4. Integral = sec⁻¹(sec x + cos x) + c.
5. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (b) \( \sec^{-1}{\left(2 x - 7 \right)} + c \)

Explanation:
1. x² - 7x + 12 = ((2x - 7)² - 1)/4, so the integrand is 2/((2x - 7)√((2x - 7)² - 1)).
2. Put 2x - 7 = u, so dx = du/2 and the integral is ∫du/(u√(u² - 1)).
3. Integral = sec⁻¹u = sec⁻¹(2x - 7) + c.
4. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) \( \frac{\tan x}1+\frac{\tan^2x}2+\cdots+\frac{\tan^9x}9 \)

Explanation:
1. I_n + I_(n-2) = tan^(n-1)x/(n - 1).
2. The expression equals (I₂ + I₀) + (I₃ + I₁) + (I₄ + I₂) + ... + (I₁₀ + I₈).
3. Each bracket is tan^(n-1)x/(n - 1) for n = 2 to 10, giving tan x/1 + tan²x/2 + ... + tan⁹x/9.
4. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Convert products and powers of trigonometric functions to sums using identities, or substitute for the simplest function.
2. Check the final form against the options by differentiation.

Answer: (c) \( \frac{x^3}3-x+2\tan^{-1}x+c \)

Explanation:
1. The condition gives f(x) = x^n + 1 (or 1 - x^n; the value f(3) = 82 selects +).
2. f(3) = 3^n + 1 = 82 gives n = 4.
3. ∫(x⁴ + 1)/(x² + 1) dx = ∫(x² - 1 + 2/(x² + 1)) dx = x³/3 - x + 2 tan⁻¹x + c.
4. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Split rational functions into partial fractions, or complete the square for quadratics.
2. Compare coefficients carefully to avoid algebra slips.

Answer: (a) \( \tan^{-1}{\left(e^{x} - e^{- x} \right)} + c \)

Explanation:
1. Divide numerator and denominator by e^(2x): (e^x + e^(-x))/(e^(2x) - 1 + e^(-2x)).
2. Put t = e^x - e^(-x), so dt = (e^x + e^(-x)) dx and e^(2x) + e^(-2x) = t² + 2.
3. ∫dt/(t² + 1) = tan⁻¹t.
4. Integral = tan⁻¹(e^x - e^(-x)).
5. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (c) \( \cos^{-1}{\left(\frac{1}{\left|{x}\right|} \right)} - \frac{\sqrt{x^{2} - 1}}{x} + c \)

Explanation:
1. Put x = sec θ (for x > 1): dx = sec θ tan θ dθ, 1/x² = cos²θ and √((x - 1)/(x + 1)) = √((1 - cos θ)/(1 + cos θ)) = tan(θ/2).
2. The integrand becomes cos θ tan θ tan(θ/2) = sin θ tan(θ/2) = 2sin²(θ/2) = 1 - cos θ.
3. ∫(1 - cos θ) dθ = θ - sin θ = cos⁻¹(1/x) - √(x² - 1)/x.
4. Integral = cos⁻¹(1/|x|) - √(x² - 1)/x + c.
5. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (b) \( (6,5) \)

Explanation:
1. (6x + 7)/(x + 2)² = 6(x + 2)/(x + 2)² - 5/(x + 2)² = 6/(x + 2) - 5/(x + 2)².
2. ∫ = 6 log|x + 2| + 5/(x + 2) + c.
3. So (A, B) = (6, 5).
4. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Split rational functions into partial fractions, or complete the square for quadratics.
2. Compare coefficients carefully to avoid algebra slips.

Answer: (c) \( 2 \log{\left(\left|{\sin^{2}{\left(x \right)} - 4 \sin{\left(x \right)} + 5}\right| \right)} + 7 \tan^{-1}{\left(\sin{\left(x \right)} - 2 \right)} + c \)

Explanation:
1. 2 sin 2x - cos x = cos x (4 sin x - 1) and 6 - cos²x - 4 sin x = sin²x - 4 sin x + 5.
2. Put sin x = t: ∫(4t - 1)/(t² - 4t + 5) dt = 2∫(2t - 4)/(t² - 4t + 5) dt + 7∫dt/((t - 2)² + 1).
3. = 2 log|t² - 4t + 5| + 7 tan⁻¹(t - 2).
4. Integral = 2 log|sin²x - 4 sin x + 5| + 7 tan⁻¹(sin x - 2) + c.
5. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (b) \( - \frac{1}{18 \left(\tan{\left(x \right)} + \sec{\left(x \right)}\right)^{9}} - \frac{1}{22 \left(\tan{\left(x \right)} + \sec{\left(x \right)}\right)^{11}} + c \)

Explanation:
1. Put sec x + tan x = t, so sec x - tan x = 1/t and sec x = (t² + 1)/(2t).
2. dt = sec x (sec x + tan x) dx, so sec²x dx = sec x dt/t = (t² + 1)/(2t²) dt.
3. The integral is (1/2)∫(t⁻¹⁰ + t⁻¹²) dt = (1/2)[-t⁻⁹/9 - t⁻¹¹/11].
4. Integral = (1/2)[-(1/9)(sec x + tan x)⁻⁹ - (1/11)(sec x + tan x)⁻¹¹] + c.
5. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. x² + 2x + 10 = (x + 1)² + 9, so ∫dx/√((x + 1)² + 3²) = sinh⁻¹((x + 1)/3). A is true.
2. R: completing the square gives a[(x + b/(2a))² + (4ac - b²)/(4a²)], and the standard form yields the stated result when 4ac - b² > 0. R is true, and with a = 1, b = 2, c = 10 it gives A directly.

Teacher's Note:
1. First decide whether the assertion and the reason are each true, then decide whether the reason actually explains the assertion.
2. Standard forms for quadratics under a root depend on the sign of the leading coefficient and of the discriminant.

Answer: (d) \( \frac\pi2x-\frac12x\cos^{-1}x+\frac12\sqrt{1-x^2}+c \)

Explanation:
1. Put x = cos θ: (1 - x)/2 = sin²(θ/2), so cos⁻¹√((1 - x)/2) = cos⁻¹(sin(θ/2)) = π/2 - θ/2 = π/2 - (1/2)cos⁻¹x.
2. ∫cos⁻¹x dx = x cos⁻¹x - √(1 - x²).
3. Integral = (π/2)x - (1/2)[x cos⁻¹x - √(1 - x²)] = (π/2)x - (1/2)x cos⁻¹x + (1/2)√(1 - x²) + c.
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.

Answer: (a) \( \left(\frac1{2a^3},\frac1{4a^3}\right) \)

Explanation:
1. 1/(a⁴ - x⁴) = (1/(2a²))[1/(a² - x²) + 1/(a² + x²)].
2. ∫dx/(a² - x²) = (1/(2a)) log|(a + x)/(a - x)| and ∫dx/(a² + x²) = (1/a) tan⁻¹(x/a).
3. So A = 1/(2a³) and B = 1/(4a³).
4. Differentiating this result returns the integrand, so option (a) is correct.

Teacher's Note:
1. Split rational functions into partial fractions, or complete the square for quadratics.
2. Compare coefficients carefully to avoid algebra slips.

Answer: (d) \( \frac{\tan^{-1}{\left(\frac{\sinh{\left(x \right)}}{2} \right)}}{2} + c \)

Explanation:
1. cosh²x + 3 = sinh²x + 4.
2. Put sinh x = t: ∫dt/(t² + 4) = (1/2) tan⁻¹(t/2).
3. Integral = (1/2) tan⁻¹(sinh x/2).
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Put log x = t, so dx/x = dt: ∫sin t dt = -cos t = -cos(log x). A is true.
2. R is the general substitution rule, of which A is an application with f(x) = log x and F = sin.

Teacher's Note:
1. First decide whether the assertion and the reason are each true, then decide whether the reason actually explains the assertion.
2. Standard forms for quadratics under a root depend on the sign of the leading coefficient and of the discriminant.

Answer: (c) \( \frac1n\sin2nx+c \)

Explanation:
1. sin²(n + 1)x - 2 sin²nx + sin²(n - 1)x = [-cos(2n + 2)x - cos(2n - 2)x + 2cos 2nx]/2 = cos 2nx(1 - cos 2x) = 2 sin²x cos 2nx.
2. So the combination is ∫2cos 2nx dx = (1/n) sin 2nx + c.
3. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Convert products and powers of trigonometric functions to sums using identities, or substitute for the simplest function.
2. Check the final form against the options by differentiation.

Answer: (d) \( \left(\frac1{2\sqrt2},-\sqrt2\right) \)

Explanation:
1. Put e^x + e^(-x) = t, so dt = (e^x - e^(-x)) dx and e^(2x) + e^(-2x) = t² - 2.
2. The integral is ∫dt/(t² - 2) = (1/(2√2)) log|(t - √2)/(t + √2)|.
3. Writing this as A log|(t + a)/(t - a)| gives a = -√2 and A = 1/(2√2).
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (a) \( \frac12\ln\left((x+y)^2+1\right)+c \)

Explanation:
1. Differentiating y(x + y)² = x gives y'(x + y)² + 2y(x + y)(1 + y') = 1, which rearranges to (1 + y')(x + y)(x + 3y) = (x + y)² + 1.
2. So d/dx ln((x + y)² + 1) = 2(x + y)(1 + y')/((x + y)² + 1) = 2/(x + 3y).
3. Hence ∫dx/(x + 3y) = (1/2) ln((x + y)² + 1) + c.

Teacher's Note:
1. Remember the standard integral formulas together with the conditions under which each holds.
2. Check every statement on its own, for example by differentiating the proposed result.

Answer: (d) \( \frac{\sqrt{x^{3} + 1} \left(2 x^{3} - 4\right)}{9} + c \)

Explanation:
1. Put 1 + x³ = t², so 3x² dx = 2t dt and x³ = t² - 1.
2. ∫x³·x²dx/√(1 + x³) = (2/3)∫(t² - 1) dt = (2/3)(t³/3 - t) = (2/9)t(t² - 3).
3. With t² = 1 + x³: (2/9)(x³ - 2)√(1 + x³).
4. Differentiating this result returns the integrand, so option (d) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (c) \( f(x)=\sin2x+1 \)

Explanation:
1. Multiply numerator and denominator by 2(cos x + sin x): the numerator becomes -2cos 2x, the denominator becomes (1 + sin 2x)√((1 + sin 2x)² - 1).
2. Put 1 + sin 2x = t, so dt = 2cos 2x dx.
3. The integral is -∫dt/(t√(t² - 1)) = -sec⁻¹t = cosec⁻¹t - π/2.
4. So f(x) = 1 + sin 2x.
5. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Choose the substitution so that the derivative of the inner function appears in the integrand.
2. After substituting, change every part of the integral to the new variable.

Answer: (b) \( \frac{x \tan^{-1}{\left(x \right)}}{2} - \frac{\log{\left(x^{2} + 1 \right)}}{4} + c \)

Explanation:
1. Put x = tan θ: (√(1 + x²) - 1)/x = (sec θ - 1)/tan θ = tan(θ/2).
2. So the integrand is θ/2 = (1/2) tan⁻¹x.
3. (1/2)∫tan⁻¹x dx = (1/2)[x tan⁻¹x - (1/2) log(1 + x²)].
4. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Use integration by parts with the ILATE order for choosing u.
2. Special forms such as the integral of e to the x times (f + f') are solved at once.