📝 Integral Tables (40 MCQs)

📖 Explanation: The table formula is a perfect match with $u = x$. The integral of $u^2 \sin u$ is correctly given as $-u^2 \cos u + 2u \sin u + 2 \cos u + C$. Students often make sign errors with the cosine term or forget the factor of 2 in the $\sin$ term, so this tests the ability to correctly apply a standard formula.

📖 Explanation: Letting $u = x-2$ gives $x = u+2$ and $dx = du$. The integral becomes $\int (u+2) \sqrt{u^2+1} \, du$. This is the correct setup to split into two integrals, $\int u \sqrt{u^2+1} \, du + 2 \int \sqrt{u^2+1} \, du$. A common mistake is forgetting to change the factor $x$ to $u+2$, leading to option B, which would be incorrect.

📖 Explanation: After integrating, we get $-\frac{1}{2} \cos \theta + C$. From the triangle for $x = \sqrt{2} \sec \theta$, $\cos \theta = \frac{\sqrt{2}}{x}$. Substituting gives $-\frac{\sqrt{2}}{2x} + C$, but this simplifies to $-\frac{\sqrt{x^2-2}}{2x} + C$ after rationalizing. Students often incorrectly invert the trigonometric ratio or fail to simplify.

📖 Explanation: The integral is $\int e^{2x} \cos x \, dx$. Comparing to $\int e^{au} \cos(bu) \, du$, we have $a=2$ and $b=1$. The formula gives $\frac{e^{2x}}{4+1}(2\cos x + \sin x) + C$. Option A is correct. A common error is to confuse the coefficients of the exponential and the trigonometric functions, leading to option B. The sign of $b$ is handled by the formula, so options C and D are also common mistakes.

📖 Explanation: The integral is a rational function in $x$. The substitution $u = 4x+3$ simplifies the denominator to a linear term, but the $x$ in the denominator remains. However, it is a standard technique to use partial fractions or a substitution like $u = 4x+3$ to reduce it to a form that matches the table entry $\int \frac{du}{u(a+bu)}$. The other substitutions are for trigonometric or radical forms, which are not appropriate here.

📖 Explanation: Substituting $m=5, n=3$ into the formula gives $\frac{\sin(2x)}{2(2)} + \frac{\sin(8x)}{2(8)} + C = \frac{\sin 2x}{4} + \frac{\sin 8x}{16} + C$. The most common mistakes are forgetting to multiply by 2 in the denominator or using the wrong sign. This tests the ability to correctly use a standard product-to-sum formula.

📖 Explanation: Let $u = x-3$. Then $x = u+3$ and $dx = du$, so the integral becomes $\int \frac{u+3}{u^2+4} du = \frac{1}{2} \ln(u^2+4) + \frac{3}{2} \tan^{-1}(u/2) + C$. Substituting back gives option A. Common errors include forgetting the constant multiplier in the $\ln$ term (option D) or in the $\tan^{-1}$ term (option C), or both (option B).

📖 Explanation: The integral $\int \sin(\ln x) \, dx$ is not a simple table match. A substitution like $u = \ln x$ transforms it to $\int e^u \sin u \, du$, which is a standard form in integral tables (exponential times sine/cosine). The others are either direct table matches or simple substitutions. This question tests the student's ability to recognize when a preliminary substitution is necessary to make an integral match a table form.

📖 Explanation: The integral $\int \tan^2 x \sec^4 x \, dx = \int \tan^2 x \sec^2 x \sec^2 x \, dx$. Let $u = \tan x$, then $\sec^2 x \, dx = du$ and $\sec^2 x = u^2 + 1$. The integral becomes $\int u^2 (u^2+1) du = \frac{u^5}{5} + \frac{u^3}{3} + C = \frac{\tan^5 x}{5} + \frac{\tan^3 x}{3} + C$. The student's result is correct. The question is designed to test the understanding of the method. Option A describes a common error but doesn't apply here.

📖 Explanation: The integral $\int_0^\pi \sin^2 x \, dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_0^\pi = \frac{\pi}{2}$. Therefore, $V = \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{2}$. A common mistake is to compute the integral as $\frac{\pi}{2}$ and forget to multiply by the factor of $\pi$ from the volume formula, leading to option B. Others may make errors in the antiderivative evaluation.

📖 Explanation: After integrating, we get $\frac{\theta}{2} + \frac{\sin 2\theta}{4} + C$. Using $\sin 2\theta = 2\sin\theta \cos\theta = 2x\sqrt{1-x^2}$, we get $\frac{1}{2}\sin^{-1}x + \frac{x\sqrt{1-x^2}}{2} + C$. Option A is correct. Option B is the same. Option C is missing the factor of $1/2$ and the inverse sine term is incorrect. This tests the ability to convert back from $\theta$ to $x$ correctly.

📖 Explanation: To match $\int \frac{du}{\sqrt{a^2 - u^2}}$, we need $a^2 = 4$ so $a=2$, and $u^2 = 9x^2$ so $u=3x$. Then $du = 3 \, dx$, and the integral becomes $\frac{1}{3} \int \frac{du}{\sqrt{4-u^2}} = \frac{1}{3} \sin^{-1}(u/2) + C$. A common error is to take $u=2x$ and $a=3$ or to not account for the $du$ factor, making option A the correct choice.

📖 Explanation: The expression $\sqrt{x^2 - a^2}$ is typically handled using the trigonometric substitution $x = a \sec \theta$ or the hyperbolic substitution $x = a \cosh u$. The substitution $x = a \sin \theta$ is used for $\sqrt{a^2 - x^2}$. Option D is incorrect because it would lead to a simplification to $\sqrt{a^2 \sin^2 \theta - a^2} = a\sqrt{-\cos^2 \theta}$, which is not a real function.

📖 Explanation: After completing the square, the integral is of the form $\int \frac{du}{u^2 + a^2}$ where $u = x-2$ and $a=2$. This matches the table formula for the arctangent. Option B is for a difference of squares, C for a reciprocal square root, and D for a reciprocal square root of the form $\sqrt{a^2-u^2}$. This tests the ability to classify integrands and select the correct table entry.

📖 Explanation: Matching the integral to the formula, we have $a=3$ and $u=x$. The antiderivative is $-\frac{1}{3} \ln\left|\frac{3 + \sqrt{x^2+9}}{x}\right| + C$. Option B is incorrect because it inverts the fraction, and option C has the wrong sign. Students often make sign errors or forget the factor of $1/a$.

📖 Explanation: Let $u = x^2$. Then $du = 2x \, dx$ and $x^3 \, dx = x^2 \cdot x \, dx = u \cdot \frac{du}{2} = \frac{u}{2} du$. The integral becomes $\frac{1}{2} \int u e^{u} \, du$, which matches the table with $a=1$. A common mistake is to not see the needed substitution or to incorrectly transform the differential. This tests the ability to combine substitution with table formulas.

📖 Explanation: The integral $\int \ln x \, dx$ is a classic example where integration by parts is used with $u = \ln x$ and $dv = dx$. The result is $x \ln x - x + C$. The table formula for $\int \ln u \, du$ is derived from integration by parts. While the others also use integration by parts, the $\int \ln x \, dx$ is the most direct and fundamental Easy. This tests the understanding of where table formulas come from.

📖 Explanation: When using numerical integration for an improper integral, the total error is the sum of the approximation error (Simpson's rule) and the truncation error from discarding the tail of the infinite interval. These errors are additive in a worst-case sense. Option A is correct. This question connects two concepts: improper integrals and numerical integration.

📖 Explanation: The integral is of the form $\int \frac{dx}{ax^2+bx+c}$. The standard technique is to complete the square to get it into the form $\int \frac{du}{u^2 + a^2}$, which matches the arctangent table formula. Completing the square gives $(x+3)^2 + 1$. The other options are not efficient or appropriate first steps. This tests the ability to recognize the need for completing the square.

📖 Explanation: The formulas are equivalent because $\sec x + \tan x = \tan(\pi/4 + x/2)$. This is a standard trigonometric identity. The constants of integration can absorb any difference. This question tests the understanding that different-looking results can be equivalent, a key concept when using tables or CAS. Option C is tempting but the identity holds for all $x$ where both are defined.

📖 Explanation: For the integral $\int \frac{dx}{x\sqrt{4-x^2}}$, we have $a=2$ and $u=x$. The correct antiderivative should be $-\frac{1}{2} \ln\left|\frac{2+\sqrt{4-x^2}}{x}\right| + C$. The student's result has a factor of $1/2$ but it should be $1/2$ not $1$. The coefficient in front of the logarithm should be $-1/2$. The student might have thought $a=1$, which is a common error when the integrand is not perfectly matched. This tests the ability to identify constants in table formulas.

📖 Explanation: Completing the square: $x^2+4x+5 = (x+2)^2 + 1$. Let $u = x+2$, then $du = dx$, and the integral becomes $\int \frac{du}{u^2+1} = \tan^{-1} u + C$. Option A is correct. Option C would introduce a factor of 2, but $u = (x+2)/2$ would make $du = dx/2$ and the denominator would be $4u^2+1$, requiring an extra factor. The simplest substitution is $u=x+2$.

📖 Explanation: With $a=2$, $\frac{1}{2a} = \frac{1}{4}$. So the antiderivative $\frac{1}{4} \ln\left|\frac{x-2}{x+2}\right| + C$ is correct. Option B is incorrect because it uses $\frac{1}{2}$, which would be for $a=1$. Option C has the fraction inverted. This tests the correct Easy of the formula and the ability to calculate $1/(2a)$.

📖 Explanation: All three identities are equivalent. $\sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x = \frac{1}{8}(1 - \cos 4x)$. The third option is also correct because $\sin^2 2x = 1 - \cos^2 2x$, so $\frac{1}{4}(1 - \cos^2 2x)$ is also valid. This question tests the ability to use and manipulate trigonometric identities to make an integral match a table form. Option D is correct because all are true identities.

📖 Explanation: Using integration by parts with $u = t^n$ and $dv = e^{-st} dt$, we get $du = n t^{n-1} dt$ and $v = -\frac{1}{s} e^{-st}$. The integral becomes $\left[-\frac{t^n}{s} e^{-st}\right]_0^\infty - \int_0^\infty \left(-\frac{1}{s} e^{-st}\right) (n t^{n-1}) dt = \left[-\frac{t^n}{s} e^{-st}\right]_0^\infty + \frac{n}{s} \int_0^\infty e^{-st} t^{n-1} dt$. Option A is correct. This is a difficult problem involving an improper integral, integration by parts, and a reduction formula.

📖 Explanation: The integral $\int \frac{x}{\sqrt{x^2+1}} \, dx$ can be solved by a simple substitution $u = x^2+1$, giving $\frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} + C = \sqrt{x^2+1} + C$. It is a direct match to a power rule and does not require the more complex logarithmic table formula. Option D is correct. A common mistake is to overcomplicate the problem and use an inappropriate table formula. This tests the ability to recognize simple integrals.

📖 Explanation: The area under $y=1/x$ from 1 to $b$ is $\ln b$. As $b \to \infty$, $\ln b \to \infty$. Therefore, the improper integral diverges. Option A is correct. Option B is incorrect because it confuses the integral with $\int_1^\infty 1/x^2 \, dx$, which converges to 1. This tests the ability to interpret graphical area as an improper integral and to understand divergence.

📖 Explanation: Here $a=2$ and $b=3$. The formula gives $\frac{1}{a} \ln\left|\frac{u}{a+bu}\right| = \frac{1}{2} \ln\left|\frac{x}{2+3x}\right| + C$. Option A is correct. Option B uses $1/b$ instead of $1/a$. Option C misses the $1/2$ factor. This tests the correct substitution of constants into a formula.

📖 Explanation: For $\sqrt{u^2 - a^2}$, the standard trigonometric substitution is $u = a \sec \theta$, leading to $\tan \theta$. For $\sqrt{u^2 + a^2}$, it's $u = a \tan \theta$. For $\sqrt{a^2 - u^2}$, it's $u = a \sin \theta$. Option A is correct. The other options are common mis-matches of substitutions to forms. This tests the foundational knowledge of trigonometric substitutions.

📖 Explanation: Completing the square: $x^2+2x+2 = (x+1)^2 + 1$. The integral becomes $\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1) + C$. Option A is correct. Option B would be for $(x+1)^2+2$. Option D is for $x^2+2x+5$. This tests the ability to complete the square and correctly use the arctangent formula.

📖 Explanation: The integral of a rational function of $\sin x$ and $\cos x$ can be tackled by the Weierstrass substitution $u = \tan(x/2)$. This transforms it into an integral of a rational function of $u$, which can be evaluated using partial fractions and table formulas. The other substitutions are not generally applicable to this form. This tests the knowledge of special substitutions for trigonometric integrals.

📖 Explanation: With $a=2$, the formula gives $\sqrt{x^2-4} - 2 \sec^{-1}(x/2) + C$. Option A is correct. Option B has the wrong coefficient for the inverse secant. Option C uses $x$ instead of $x/2$. Option D has the wrong sign. This tests the correct Easy of a formula and the substitution of constants.

📖 Explanation: Let $u = x+1$, then $du = dx$. The integral becomes $\int \ln u \, du = u \ln u - u + C = (x+1)\ln(x+1) - (x+1) + C$. Option A is correct. Option B is a common mistake where the student incorrectly integrates without changing the variable. This tests the understanding of the substitution in a table formula.

📖 Explanation: The integral $\int e^{x^2} \, dx$ is a classic example of a non-elementary integral. It cannot be expressed in terms of polynomials, exponentials, logarithms, trigonometric functions, etc. The error function is a special function defined for this purpose. This question tests the understanding of the limitations of elementary functions and the role of CAS and tables. Option A is correct.

📖 Explanation: The integral $\int \sqrt{1+e^x} \, dx$ becomes $\int \frac{\sqrt{1+u}}{u} du$ after $u = e^x$. This is a standard form in tables (e.g., $\int \frac{\sqrt{a+bu}}{u} du$). The others are either direct matches, simple substitutions, or standard forms. This tests the ability to recognize when a substitution will bring an integral into a table form.

📖 Explanation: Let $x = 3\sin \theta$. Then $dx = 3\cos \theta \, d\theta$ and $\sqrt{9-x^2} = 3\cos \theta$. The integral becomes $\int \frac{3\cos \theta \, d\theta}{9\sin^2 \theta (3\cos \theta)} = \frac{1}{9} \int \csc^2 \theta \, d\theta = -\frac{1}{9} \cot \theta + C = -\frac{1}{9} \frac{\cos \theta}{\sin \theta} + C = -\frac{\sqrt{9-x^2}}{9x} + C$. Option A is correct. This tests multiple steps: substitution, integration, and back-substitution.