JEE Mathematics Indefinite Integration MCQs Set 14

Answer: (b) \( \frac{2 \sqrt{x}}{\sqrt[4]{x^{2} + 1}} + c \)

Explanation:
1. Write (1 + x²)^(5/4) = x^(5/2)(1 + x⁻²)^(5/4), so the integrand is x⁻³(1 + x⁻²)^(-5/4).
2. Put 1 + x⁻² = t, so -2x⁻³ dx = dt.
3. Integral = -(1/2)∫t^(-5/4) dt = 2t^(-1/4) = 2√x/(1 + x²)^(1/4).
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) \( \frac{e^{x} \cos{\left(x \right)}}{\sin{\left(x \right)} + 1} + c \)

Explanation:
1. Take f = cos x/(1 + sin x); then f' = -1/(1 + sin x), so f + f' = (cos x - 1)/(1 + sin x).
2. The integral of e^x(f + f') is e^x f.
3. Result = e^x cos x/(1 + sin 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: (a) \( - x + \frac{4 x \sin^{-1}{\left(x \right)} + 4 \sqrt{1 - x^{2}}}{\pi} + c \)

Explanation:
1. Since sin⁻¹x + cos⁻¹x = π/2 and cos⁻¹x = π/2 - sin⁻¹x, the integrand is (4/π) sin⁻¹x - 1.
2. ∫sin⁻¹x dx = x sin⁻¹x + √(1 - x²).
3. Result = (4/π)[x sin⁻¹x + √(1 - x²)] - 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: (a) \( \frac{\log{\left(\left|{\frac{\tan{\left(x \right)}}{\tan{\left(x \right)} + 2}}\right| \right)}}{2} + c \)

Explanation:
1. Divide numerator and denominator by cos²x: ∫sec²x dx/(tan²x + 2 tan x).
2. Put tan x = t: ∫dt/(t(t + 2)) = (1/2) log|t/(t + 2)|.
3. Result = (1/2) log|tan x/(tan x + 2)|.
4. 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: (a) \( \left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x} + c \)

Explanation:
1. 2^(3 log₂x) = x³, so the integral is ∫x³e^x dx.
2. Using ∫e^x f = e^x[f - f' + f'' - f'''] with f = x³: e^x(x³ - 3x² + 6x - 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) \( - \frac{\cos{\left(4 x \right)}}{8} + c \)

Explanation:
1. cos 4x + 1 = 2cos²2x and cot x - tan x = 2 cot 2x.
2. The integrand is cos²2x tan 2x = sin 2x cos 2x = (1/2) sin 4x.
3. Integral = -(1/8) cos 4x, so A = -1/8.
4. 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: (c) \( \frac{x-1}{3} \)

Explanation:
1. Write 2x + 5 = 2(x - 1) + 7.
2. ∫2(x - 1)/√(x² - 2x + 10) dx = 2√(x² - 2x + 10).
3. ∫7 dx/√((x - 1)² + 9) = 7 sinh⁻¹((x - 1)/3).
4. So f(x) = (x - 1)/3.
5. 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: (c) \( 2 \log{\left(\left|{\sec{\left(\frac{x}{2} \right)}}\right| \right)} + c \)

Explanation:
1. tan(A - B) = (tan A - tan B)/(1 + tan A tan B), so the integrand is tan(x - x/2) = tan(x/2).
2. ∫tan(x/2) dx = -2 log|cos(x/2)| = 2 log|sec(x/2)|.
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) \( \left(x + \sqrt{x^{2} + 1}\right) e^{x} + c \)

Explanation:
1. Take f = x + √(1 + x²); then f' = 1 + x/√(1 + x²) = f/√(1 + x²).
2. So f + f' = f(1 + 1/√(1 + x²)), which is the integrand's bracket.
3. Result = e^x(x + √(x² + 1)).
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) \( \log{\left(\left|{\frac{2 \log{\left(x \right)} + 1}{3 \log{\left(x \right)} + 2}}\right| \right)} + c \)

Explanation:
1. Put log x = t, so dx/x = dt.
2. 6t² + 7t + 2 = (2t + 1)(3t + 2), and 1/((2t + 1)(3t + 2)) = 2/(2t + 1) - 3/(3t + 2).
3. Integral = log|(2t + 1)/(3t + 2)|.
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: (c) \( \tan^{-1}{\left(\sin{\left(x \right)} \right)} + c \)

Explanation:
1. Put sin x = t, so cos x dx = dt.
2. The integral is ∫dt/(1 + t²) = tan⁻¹t.
3. Result = tan⁻¹(sin x).
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: (b) \( \frac{\log{\left(\sin^{8}{\left(x \right)} \left|{\frac{1}{\sin^{9}{\left(x \right)} + \cos^{9}{\left(x \right)}}}\right| \left|{\sin{\left(x \right)}}\right| \right)}}{9} + c \)

Explanation:
1. Divide numerator and denominator by cos²x: ∫sec²x dx/(tan x (tan⁹x + 1)).
2. Put tan x = t: ∫dt/(t(t⁹ + 1)) = (1/9) log|t⁹/(t⁹ + 1)|.
3. Result = (1/9) log|sin⁹x/(sin⁹x + cos⁹x)|.
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) \( \frac{x^{n + 1} \left(\log{\left(x \right)} - \frac{1}{n + 1}\right)}{n + 1} + c \)

Explanation:
1. By parts with u = log x and v' = x^n: ∫x^n log x dx = x^(n+1) log x/(n + 1) - ∫x^n/(n + 1) dx.
2. = x^(n+1) log x/(n + 1) - x^(n+1)/(n + 1)².
3. Result = [x^(n+1)/(n + 1)][log x - 1/(n + 1)].
4. 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: (a) \( \frac{2 \cos^{\frac{7}{2}}{\left(x \right)}}{7} - \frac{2 \cos^{\frac{3}{2}}{\left(x \right)}}{3} + c \)

Explanation:
1. Write sin³x = sin x(1 - cos²x) and put cos x = t, so sin x dx = -dt.
2. The integral is -∫√t(1 - t²) dt = -(2/3)t^(3/2) + (2/7)t^(7/2).
3. Result = -(2/3)cos^(3/2)x + (2/7)cos^(7/2)x.
4. 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) \( \frac{2}{n-1}\sin(n-1)x-I_{n-2} \)

Explanation:
1. cos nx + cos(n - 2)x = 2 cos(n - 1)x cos x.
2. So cos nx/cos x = 2 cos(n - 1)x - cos(n - 2)x/cos x.
3. Integrating, I_n = (2/(n - 1)) sin(n - 1)x - I_(n-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: (c) \( \log(1+\sqrt2)-\frac\pi4 \)

Explanation:
1. Rationalise: x²/(1 + √(1 + x²)) = √(1 + x²) - 1.
2. So the integrand is (√(1 + x²) - 1)/(1 + x²) = 1/√(1 + x²) - 1/(1 + x²).
3. f(x) = sinh⁻¹x - tan⁻¹x, and f(1) = log(1 + √2) - π/4.
4. 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: (b) \( \frac{x^{2} \sin{\left(x^{2} \right)}}{2} + \frac{\cos{\left(x^{2} \right)}}{2} + c \)

Explanation:
1. Put x² = t, so x dx = dt/2 and the integral is (1/2)∫t cos t dt.
2. By parts, ∫t cos t dt = t sin t + cos t.
3. Result = (1/2)[x² sin x² + cos 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.

Answer: (a) \( \frac{3^{x} \left(4 \sin{\left(4 x \right)} + \log{\left(3 \right)} \cos{\left(4 x \right)}\right)}{\log{\left(3 \right)}^{2} + 16} + c \)

Explanation:
1. 3^x = e^(x log 3), so use ∫e^(ax) cos bx dx = e^(ax)(a cos bx + b sin bx)/(a² + b²) with a = log 3, b = 4.
2. Result = 3^x[(log 3) cos 4x + 4 sin 4x]/[(log 3)² + 16].
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: (a) \( \frac{- \frac{a^{2}}{a + b x} - 2 a \log{\left(a + b x \right)} + a + b x}{b^{3}} + c \)

Explanation:
1. Put a + bx = t, so x = (t - a)/b and dx = dt/b.
2. The integral is (1/b³)∫(t - a)²/t² dt = (1/b³)∫(1 - 2a/t + a²/t²) dt.
3. Result = (1/b³)[(a + bx) - 2a log(a + bx) - a²/(a + bx)].
4. 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: (a) \( \frac12\left\{x^2g(x^2)-\int g(x^2)\,d(x^2)\right\} \)

Explanation:
1. Put x² = t, so x dx = dt/2 and the integral is (1/2)∫t f(t) dt.
2. By parts: ∫t f(t) dt = t g(t) - ∫g(t) dt.
3. So the integral is (1/2){x²g(x²) - ∫g(x²) d(x²)}.
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: (c) \( 2 \tan^{-1}{\left(\sqrt{x + 99} \right)} + c \)

Explanation:
1. Put √(x + 99) = t, so dx = 2t dt and x + 100 = t² + 1.
2. The integral is 2∫dt/(1 + t²) = 2 tan⁻¹t.
3. Result = 2 tan⁻¹√(x + 99).
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: (b) \( \frac{\log{\left(\left|{x^{e} + e^{x}}\right| \right)}}{e} + c \)

Explanation:
1. Note d/dx(e^x + x^e) = e^x + e x^(e-1), while the numerator is (1/e)(e^x + e x^(e-1)).
2. The integral is (1/e) log|e^x + x^e|.
3. Differentiating this result returns the integrand, so option (b) is correct.

Teacher's Note:
1. Learn the standard integrals thoroughly; most one-line questions reduce to them after a small rewrite.
2. Always verify by differentiating the answer when options look similar.