JEE Mathematics Indefinite Integration MCQs Set 19

Answer: (b) \( \frac{\left(x + \sqrt{x^{2} + 2}\right)^{\frac{3}{2}}}{3} - \frac{2}{\sqrt{x + \sqrt{x^{2} + 2}}} + c \)

Explanation:
1. Put x + √(x² + 2) = t. Then x² + 2 = (t - x)² gives x = (1/2)(t - 2/t) and dx = (1/2)(1 + 2/t²) dt.
2. The integral is (1/2)∫√t (1 + 2/t²) dt = (1/2)∫(t^(1/2) + 2t^(-3/2)) dt = (1/2)[(2/3)t^(3/2) - 4t^(-1/2)].
3. Integral = (1/3)(x + √(x² + 2))^(3/2) - 2(x + √(x² + 2))^(-1/2) + 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: (c) \( \frac{x^{2} e^{x^{2}} e^{x^{4}}}{2} + c \)

Explanation:
1. Put x² = t, so x dx = dt/2 and the integral is (1/2)∫e^(t² + t)(1 + t + 2t²) dt.
2. Note that d/dt[t e^(t² + t)] = e^(t² + t)(1 + t(2t + 1)) = e^(t² + t)(1 + t + 2t²).
3. The integral is (1/2) t e^(t² + t).
4. Integral = (1/2) x² e^(x²) e^(x⁴) + c.
5. Differentiating this result returns the integrand, so option (c) 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: (d) \( \left(-1,\frac1{10},\frac1{10}\right) \)

Explanation:
1. The denominator is x(x¹⁰ - 1). Put x⁵ = t: dx/x = dt/(5t), so the integral is (1/5)∫dt/(t(t² - 1)).
2. 1/(t(t² - 1)) = -1/t + (1/2)/(t - 1) + (1/2)/(t + 1).
3. The integral is (1/5)[-log t + (1/2) log(t - 1) + (1/2) log(t + 1)] = -log x + (1/10) log(x⁵ - 1) + (1/10) log(x⁵ + 1).
4. So (A, B, C) = (-1, 1/10, 1/10).
5. Differentiating this result returns the integrand, so option (d) 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) \( \frac1{\sqrt2} \)

Explanation:
1. sin x + cos x = √2 sin(x + π/4).
2. ∫dx/(√2 sin(x + π/4)) = (1/√2)∫cosec(x + π/4) dx = (1/√2) log|tan(x/2 + π/8)|.
3. So k = 1/√2.
4. 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: (b) \( - \frac{x^{2}}{2} + x \tan{\left(x \right)} - \log{\left(\left|{\sec{\left(x \right)}}\right| \right)} + c \)

Explanation:
1. (sec 2x - 1)/(sec 2x + 1) = (1 - cos 2x)/(1 + cos 2x) = tan²x.
2. ∫x tan²x dx = ∫x(sec²x - 1) dx = x tan x - ∫tan x dx - x²/2.
3. Integral = x tan x - log|sec x| - x²/2.
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.

Answer: (a) \( \frac{x}{\log{\left(x \right)}} + c \)

Explanation:
1. log(x/e) = log x - 1, so the integrand is (log x - 1)/(log x)².
2. This is d/dx [x/log x].
3. Integral = x/log x.
4. 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: (d) \( \frac{\tan{\left(x \right)}}{2} + \frac{\sqrt{2} \tan^{-1}{\left(\sqrt{2} \tan{\left(x \right)} \right)}}{4} + c \)

Explanation:
1. 1 - sin⁴x = (1 - sin²x)(1 + sin²x) = cos²x (1 + sin²x).
2. Divide numerator and denominator by cos⁴x: sec⁴x/(sec²x + tan²x) and put tan x = t: (1 + t²)/(1 + 2t²) dt.
3. (1 + t²)/(1 + 2t²) = (1/2)[1 + 1/(1 + 2t²)], so the integral is (1/2)[t + (1/√2) tan⁻¹(√2 t)].
4. Integral = (1/2)[tan x + (1/√2) tan⁻¹(√2 tan x)] + c.
5. Differentiating this result returns the integrand, so option (d) 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) \( \frac12\sin^{-1}\left(\frac ta\right)+\log\sqrt{t+\sqrt{a^2-t^2}}+k \)

Explanation:
1. Put t = a sin θ: the integral is ∫cos θ dθ/(sin θ + cos θ).
2. cos θ = (1/2)[(cos θ + sin θ) + (cos θ - sin θ)], so the integral is (1/2)θ + (1/2) log|sin θ + cos θ|.
3. = (1/2) sin⁻¹(t/a) + (1/2) log((t + √(a² - t²))/a) = (1/2) sin⁻¹(t/a) + log√(t + √(a² - t²)) + k.
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. (1 + x log x)/x = log x + 1/x, which is f(x) + f'(x) for f(x) = log x.
2. By R the integral is e^x log x + c. So A is true and R explains it.

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

Explanation:
1. x^(1/n) - x^(1/(n+1)) = x^(1/(n+1)) (x^(1/(n(n+1))) - 1).
2. n² x^(1/(n+1)) (x^(1/(n(n+1))) - 1) → 1 · log x, since n² · (log x)/(n(n + 1)) → log x.
3. So f(x) = log x.
4. ∫x log x dx = (x²/2) log x - x²/4 + c.
5. 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.

Answer: (c) \( - \frac{7}{4 \tan^{\frac{4}{7}}{\left(x \right)}} + c \)

Explanation:
1. Write the integrand as tan^(-11/7)x · sec²x (since cos^(-3/7) sin^(-11/7) = sin^(-11/7) cos^(-11/7) cos^(8/7) ... = tan^(-11/7) x sec²x).
2. Put tan x = t: ∫t^(-11/7) dt = -(7/4) t^(-4/7).
3. Integral = -(7/4) tan^(-4/7)x + c.
4. 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: (a) \( x(\log x)^6 \)

Explanation:
1. By parts: I_n = x(log x)^n - n I_(n-1).
2. So I₆ + 6I₅ = x(log x)⁶.
3. 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: (b) \( \frac{- 2 \sin{\left(\log{\left(x \right)} \right)} - \cos{\left(\log{\left(x \right)} \right)}}{5 x^{2}} + c \)

Explanation:
1. Put log x = t, so x = e^t and dx/x³ = e^(-2t) dt.
2. ∫e^(-2t) sin t dt = e^(-2t)(-2 sin t - cos t)/5.
3. Integral = -(1/(5x²))[2 sin(log x) + cos(log x)].
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.

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

Explanation:
1. Divide by cos²x... the integrand is cos²x/(1 + sin²x); with t = tan x it becomes dt/((1 + t²)(1 + 2t²)).
2. 1/((1 + t²)(1 + 2t²)) = 2/(1 + 2t²) - 1/(1 + t²).
3. Integral = √2 tan⁻¹(√2 tan x) - tan⁻¹(tan x).
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: (b) \( 3 \log{\left(\tan{\left(x \right)} \right)} + \frac{\tan^{4}{\left(x \right)}}{4} + \frac{3 \tan^{2}{\left(x \right)}}{2} - \frac{\cot^{2}{\left(x \right)}}{2} + c \)

Explanation:
1. Divide numerator and denominator by cos⁸x: sec⁸x/tan³x.
2. With tan x = t, sec⁸x dx = (1 + t²)³ dt, so the integral is ∫(1 + t²)³/t³ dt = ∫(t⁻³ + 3t⁻¹ + 3t + t³) dt.
3. = -1/(2t²) + 3 log t + (3/2)t² + t⁴/4.
4. Integral = 3 log tan x + (3/2)tan²x + (1/4)tan⁴x - (1/2)cot²x + c.
5. Differentiating this result returns the integrand, so option (b) 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: (a) \( 2\sqrt2 \)

Explanation:
1. Divide numerator and denominator by x²: (1 + 4/x²)dx/(x² + 16/x²).
2. Put x - 4/x = t, so dt = (1 + 4/x²) dx and x² + 16/x² = t² + 8.
3. ∫dt/(t² + 8) = (1/(2√2)) tan⁻¹(t/(2√2)) = (1/(2√2)) tan⁻¹((x² - 4)/(2√2 x)).
4. So k = 2√2.
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) \( \frac12e^{\sec x^2}+C \)

Explanation:
1. Write the integrand as x tan x² sec x² e^(sec x²).
2. d(sec x²) = 2x sec x² tan x² dx, so the integral is (1/2)∫e^(sec x²) d(sec x²).
3. = (1/2) e^(sec x²) + 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: (b) \( \frac{x^{m+1}}{(\log x)^n} \)

Explanation:
1. Integrate by parts with u = (log x)^(-n) and dv = x^m dx.
2. I_(m,n) = x^(m+1)/((m + 1)(log x)^n) + (n/(m + 1)) ∫x^m/(log x)^(n+1) dx.
3. Multiply by (m + 1): (m + 1)I_(m,n) - n I_(m,n+1) = x^(m+1)/(log x)^n.
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.

Answer: (c) \( \frac{(1-x)^{13}}{13}-\frac{(1-x)^{12}}{12}+c \)

Explanation:
1. By the binomial theorem the bracket is x(1 - x)^11.
2. Put 1 - x = t: ∫x(1 - x)^11 dx = -∫(1 - t)t^11 dt = -t^12/12 + t^13/13.
3. Integral = (1 - x)^13/13 - (1 - x)^12/12 + c.
4. 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: (d) \( - \frac{x}{2} + \frac{\sin{\left(x \right)}}{2} - \frac{\cos{\left(x \right)}}{2} + c \)

Explanation:
1. The denominator is (sin x + cos x + 1)/(sin x cos x).
2. So the integrand is sin x cos x/(1 + sin x + cos x).
3. Since (sin x + cos x + 1)(sin x + cos x - 1) = 2 sin x cos x, the integrand equals (1/2)(sin x + cos x - 1).
4. Integral = (1/2)(sin x - cos x - x) + k.
5. Differentiating this result returns the integrand, so option (d) 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) 15

Explanation:
1. Put 1 + 2x³ = x³t³ (so t³ = 2 + x⁻³). Then 3t² dt = -3x⁻⁴ dx, so dx = -x⁴t² dt.
2. The integral becomes -∫x⁻⁶ (x³t³)^(2/3) x⁴ t² dt = -∫t⁴ dt = -t⁵/5 + c.
3. t⁵ = (2 + 1/x³)^(5/3), so the integral is -(1/5)(2 + 1/x³)^(5/3) + c.
4. K = 5, L = 2, M = 5, N = 3, so K + L + M + N = 15.

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: (c) \( \frac{x^{2}}{2} - x + c \)

Explanation:
1. x² - √x = √x(x√x·... ) = √x(x^(3/2) - 1) = √x(√x - 1)(x + √x + 1).
2. The denominator is √x(x + √x + 1).
3. The integrand reduces to (√x + 1)(√x - 1) = x - 1.
4. Integral = x²/2 - x.
5. Differentiating this result returns the integrand, so option (c) is correct.

Teacher's Note:
1. Rationalise or simplify the integrand algebraically before integrating.
2. Substitution of a power of x often removes fractional exponents.

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

Explanation:
1. 2x - 3x² + 1 = 3[4/9 - (x - 1/3)²], so the integral is (1/√3) sin⁻¹((x - 1/3)/(2/3)) = (1/√3) sin⁻¹((3x - 1)/2). A is true.
2. R is the standard form for a < 0 and a positive discriminant; with a = -3, b = 2, c = 1 it gives (6x - 2)/4 = (3x - 1)/2, which is A.

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.