📝 Basic integration formulas list (30 MCQs)

📖 Explanation: The power rule for integration, $\int x^r \, dx = \frac{x^{r+1}}{r+1} + C$, fails when $r = -1$ because it results in division by zero. This is the foundational reason why the integral of $1/x$ is a special case, resulting in the natural logarithm function $\ln|x| + C$. The student's attempt to apply the power rule in this case shows a lack of understanding of its limitations and the exceptional nature of the reciprocal function.

📖 Explanation: The derivative of $\sec x$ is $\sec x \tan x$. Therefore, the antiderivative of $\sec x \tan x$ is $\sec x + C$. The student's answer, $\ln|\sec x| + C$, is the antiderivative of $\tan x$. This is a Easy question testing the knowledge of standard trigonometric derivatives and their corresponding integration formulas, which are fundamental to integral calculus.

📖 Explanation: The integration formula is the inverse operation of differentiation. If the derivative of $\tan^{-1} x$ is $1/(1+x^2)$, then the integral of $1/(1+x^2)$ is $\tan^{-1} x$. The constant of integration, $C$, is essential to represent the family of all antiderivatives. This question tests the fundamental relationship between differentiation and integration, a core concept in calculus.

📖 Explanation: The derivative of $\sin^{-1} x$ is $1/\sqrt{1-x^2}$, and the derivative of $-\cos^{-1} x$ is also $1/\sqrt{1-x^2}$. Since antiderivatives can differ by a constant, both $\sin^{-1} x + C$ and $-\cos^{-1} x + C$ are correct. This highlights the fact that antiderivatives are not unique and can be expressed in multiple ways due to trigonometric identities, which is a key Medium in integration.

📖 Explanation: Using the constant multiple rule and the basic integral of cosine, $\int 4 \cos x \, dx = 4 \int \cos x \, dx = 4 \sin x + C$. The constant $C$ is a single arbitrary constant. Option C is incorrect because it introduces an unnecessary constant $4C$, which is still just an arbitrary constant and should be simplified. This question assesses the ability to apply a standard formula and the property of constants in integration.

📖 Explanation: The sum/difference rule for indefinite integrals states that the integral of a sum is the sum of the integrals. This is a direct consequence of the sum rule for derivatives. The student's statement is a correct Easy of this property. The correct evaluation is $\frac{x^3}{3} + x + C$. This question tests the understanding of the linearity of integration, a fundamental property.

📖 Explanation: The function $1/\sqrt{x} = x^{-1/2}$. Applying the power rule, $\int x^{-1/2} \, dx = \frac{x^{1/2}}{1/2} + C = 2x^{1/2} + C = 2\sqrt{x} + C$. The student incorrectly added 1 to the exponent $-1/2$ to get $1/2$ but then used the derivative rule instead of the integration rule. This is a classic error in applying the power rule for integration.

📖 Explanation: This integral is a perfect candidate for $u$-substitution. Let $u = x^2$, then $du = 2x \, dx$. The integral becomes $\frac{1}{2} \int \sin u \, du = -\frac{1}{2} \cos u + C = -\frac{1}{2} \cos(x^2) + C$. Recognizing the composition of functions where the derivative of the inner function is present is the skill being tested here, linking substitution to standard formulas.

📖 Explanation: This is a multi-step problem. First, simplify the integrand by dividing each term in the numerator by $x^2$, giving $1 + x^{-2}$. Then apply the power rule: $\int 1 \, dx + \int x^{-2} \, dx = x - \frac{1}{x} + C$. Option A is correct. This tests the ability to manipulate an algebraic expression to make it integrable using basic formulas, a crucial skill in calculus.

📖 Explanation: We need to find the antiderivative of each term. The antiderivative of $\sec^2 x$ is $\tan x$. The antiderivative of $\csc x \cot x$ is $-\csc x$. Combining them gives $\tan x - \csc x + C$. This question tests the student's ability to recognize derivatives of trigonometric functions in reverse, which is the basis of integration, and to apply the sum rule.

📖 Explanation: Since f'(x) = 3, the antiderivative is $f(x) = 3x + C$, which represents a family of straight lines with a slope of 3. The graph of any antiderivative is a line. This question connects a graphical representation of a derivative to the algebraic form of its antiderivative. The constant $C$ corresponds to the vertical shift of the line, but the slope remains constant at 3.

📖 Explanation: The natural logarithm function $\ln x$ is only defined for positive real numbers. The integral $\int \frac{1}{x} \, dx$ requires the absolute value to extend the domain to $x < 0$. The derivative of $\ln|x|$ is $1/x$ for all $x \neq 0$. Option B is the correct response. This is a classic and important subtlety in integral calculus that students often overlook.

📖 Explanation: Position is the antiderivative of velocity. $\int \cos t \, dt = \sin t + C$. Using the initial condition $s(0) = 0$, we have $0 = \sin(0) + C$, so $C = 0$. Therefore, $s(t) = \sin t$. This applies the integral of cosine to solve a classic physics problem, linking calculus to rectilinear motion.

📖 Explanation: The derivative of both $\tan^{-1} x$ and $-\tan^{-1}(1/x)$ is $1/(1+x^2)$ (for $x \neq 0$). Although they differ by a constant on their intervals of definition, they are antiderivatives of the same function. This question highlights that antiderivatives are not unique and can have different algebraic forms due to trigonometric identities, reinforcing the concept of the constant of integration.

📖 Explanation: The antiderivative of $1/x$ is $\ln|x|$. Evaluating from 1 to 2 gives $\ln|2| - \ln|1| = \ln 2 - 0 = \ln 2$. This is a straightforward Easy of the Fundamental Theorem of Calculus after finding the antiderivative. It tests the student's ability to connect indefinite integration formulas to definite integral evaluation.

📖 Explanation: The correct formula is $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. Here $a = 2$, so the result is $\frac{1}{2} \tan^{-1}\left(\frac{x}{2}\right) + C$. The student forgot the constant factor $1/a$. This is a common error when using generalized integration formulas, testing the student's ability to apply and adjust standard formulas correctly.

📖 Explanation: Rewrite $\frac{\cos x}{\sin^2 x} = \frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = \csc x \cot x$. The integral of $\csc x \cot x$ is a standard formula. This question requires simplifying a trigonometric expression and recognizing a standard derivative-in-reverse (of $\csc x$). It tests Medium of trigonometric identities and their relation to integration formulas.

📖 Explanation: The constant multiple rule allows us to write $\int 5e^x \, dx = 5 \int e^x \, dx = 5e^x + C$. The constant $C$ is an arbitrary constant of integration. While $5e^x + 5C$ is also technically correct, it is conventionally simplified to $5e^x + C$. This question tests the understanding of the constant multiple rule and the meaning of the constant of integration. Option B is a common mistake where students think constants multiply within the function.

📖 Explanation: Displacement is the antiderivative of velocity. $\int \frac{1}{1+t^2} \, dt = \tan^{-1} t + C$. With $s(0) = 0$, $C = 0$. Thus $s(t) = \tan^{-1} t$. By the Fundamental Theorem of Calculus, $s(t) = \int_0^t v(u) \, du$, which is the area under the velocity curve from 0 to $t$. This directly links the integration formula for $1/(1+x^2)$ to a physical interpretation.

📖 Explanation: The Fundamental Theorem of Calculus states that if $F(x) = \int f(x) \, dx$, then F&#039;(x) = f(x). Since $\int \sec x \tan x \, dx = \sec x + C$, its derivative is $\sec x \tan x$. The constant $C$ disappears upon differentiation. This question tests the inverse relationship between differentiation and integration, a core concept.

📖 Explanation: The slope is $dy/dx = 3x^2$. Integrating gives $y = x^3 + C$. Using the point $(1, 2)$: $2 = 1^3 + C$, so $C = 1$. Thus $y = x^3 + 1$. This applies integration formulas to solve a geometric problem, connecting calculus to curves and slopes.

📖 Explanation: Let $u = x^2+1$, then $du = 2x \, dx$. So $\int \frac{x}{x^2+1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln(x^2+1) + C$. The student forgot the factor $1/2$ from the chain rule. This is a classic $u$-substitution error, testing the ability to apply the technique correctly.

📖 Explanation: Using the generalized formula $\int \frac{du}{u\sqrt{u^2 - a^2}} = \frac{1}{a} \sec^{-1}\left|\frac{u}{a}\right| + C$. Here $u = x$ and $a = 2$, so the integral is $\frac{1}{2} \sec^{-1}(x/2) + C$. This tests the ability to recall and apply a more advanced integration formula, including the constant factor $1/a$, which is a common point of error.

📖 Explanation: All antiderivatives of $\sin x$ are of the form $-\cos x + C$. Options A, B, and C are all of this form (with $C = 5, -\pi, 1$, respectively). Option D, $\cos x$, is not an antiderivative of $\sin x$ because its derivative is $-\sin x$, not $\sin x$. This question tests the understanding of the family of antiderivatives and the requirement that the derivative of the candidate must equal the integrand.

📖 Explanation: Integrating acceleration gives velocity: $v(t) = \int 10 \, dt = 10t + C$. Using $v(0) = 5$, we get $C = 5$, so $v(t) = 10t + 5$. This is a direct Easy of the constant multiple rule and the integral of a constant, linking it to a physical context.

📖 Explanation: The student is confusing integration with differentiation or just taking the reciprocal. The correct formula for $\int \frac{dx}{x^2 + a^2}$ is $\frac{1}{a} \tan^{-1}(x/a) + C$. Here $a = 2$, so the answer is $\frac{1}{2} \tan^{-1}(x/2) + C$. This is a profound misunderstanding of what an integral is; the student is not performing the inverse operation of differentiation.

📖 Explanation: If f&#039;(x) is a line with a positive slope, then its derivative f&#039;&#039;(x) is a positive constant. A positive second derivative indicates that the function $f(x)$ is concave up. This question links the graphical behavior of a derivative to the concavity of its antiderivative, a key concept in curve sketching and analysis.

📖 Explanation: The student's understanding is incomplete. While the derivative of $\ln x$ is $1/x$ for $x > 0$, the function $1/x$ is defined for all $x \neq 0$. The antiderivative must be valid for all $x \neq 0$. The function $\ln|x|$ satisfies this because its derivative is $1/x$ for both positive and negative $x$. This highlights the importance of domain when finding antiderivatives. The constant of integration $C$ is also crucial, as it represents the entire family of functions whose derivative is $1/x$.

📖 Explanation: This integral can be solved using two different substitutions. If $u = \sec x$, $du = \sec x \tan x \, dx$, the integral becomes $\int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2} \sec^2 x + C$. If $u = \tan x$, $du = \sec^2 x \, dx$, the integral becomes $\int u \, du = \frac{1}{2} u^2 + C = \frac{1}{2} \tan^2 x + C$. Since $\frac{1}{2} \sec^2 x = \frac{1}{2} (\tan^2 x + 1)$, the two results differ by a constant, and both are valid antiderivatives. This question tests the student's ability to see multiple solution paths and understand that antiderivatives are families of functions.