📝 U substitution for nonlinear functions (25 MCQs)

📖 Explanation: The correct substitution is $u = 2x^3 + 5$, so $du = 6x^2 \, dx$. The integral becomes $\int 2u^4 \, du = \frac{2u^5}{5} + C = \frac{2}{5}(2x^3+5)^5 + C$. Option A is missing the factor of 2, option B incorrectly divides by 5 instead of $\frac{2}{5}$, and option D fails to account for the chain rule.

📖 Explanation: The presence of $x^3$ and $e^{x^4}$ suggests that the derivative of $x^4$ is $4x^3$, which is present in the integrand up to a constant factor. Letting $u = x^4$ makes $du = 4x^3 dx$, perfectly matching the $x^3 dx$ part. The other options either don't simplify the exponential argument or don't produce $du$ in the integrand.

📖 Explanation: Using $u = 3x^2+1$, $du = 6x dx$. When $x=0, u=1$; when $x=1, u=4$. The integral becomes $\int_1^4 u^2 du = \frac{u^3}{3} \big|_1^4 = \frac{64-1}{3} = 21$. Option A is 14, which is the correct answer if one mistakenly evaluates the integral without changing limits, a common error. This question tests the understanding of changing limits for definite integrals.

📖 Explanation: If $u = \sin x$, then $du = \cos x dx$. The $\cos x$ in the integrand is exactly $du$. Therefore the integral simplifies directly to $\int u^2 du$. Options B and C incorrectly leave $\cos x$ in the expression; Option D is the result of an incorrect derivation, not an Easy of chain rule.

📖 Explanation: Let $u = 4-x^2$, $du = -2x dx$. Then $x dx = -\frac{1}{2} du$. The integral becomes $-\frac{1}{2} \int u^{-1/2} du = -\sqrt{u} + C = -\sqrt{4-x^2} + C$. Option A misses the negative sign introduced by the derivative. This is a classic case where students forget to account for the negative sign from the chain rule during substitution.

📖 Explanation: Rewriting the integrand as $\int \frac{\sin x}{\cos x} dx$, we let $u = \cos x$, $du = -\sin x dx$. The integral becomes $-\int \frac{du}{u} = -\ln|u| + C = -\ln|\cos x| + C$. Option B is a common mistake from using $u = \sin x$. Option C is equivalent to A only if you ignore the constant, but option A is the most direct standard form.

📖 Explanation: Let $u = 2x+1$, so $x = \frac{u-1}{2}$, $dx = \frac{1}{2}du$. The integral becomes $\frac{1}{4} \int (u-1)u^{1/2} du = \frac{1}{4} \int (u^{3/2} - u^{1/2}) du = \frac{(2x+1)^{5/2}}{10} - \frac{(2x+1)^{3/2}}{6} + C$. Option A is the only one that correctly handles both the $x$ and $dx$ transformations, highlighting the need to express the entire integrand in terms of $u$.

📖 Explanation: For a substitution to work, the derivative of $u$ must appear in the integrand. If $u = x^2$, $du = 2x dx$, leaving $e^{x^3}$ which cannot be expressed in terms of $u$. The correct substitution is $u = x^3$, because its derivative $3x^2 dx$ is present. Option A correctly identifies the failure of this substitution.

📖 Explanation: Let $u = e^x+1$, then $du = e^x dx$. The integral becomes $\int \frac{du}{u^2} = -\frac{1}{u} + C = -\frac{1}{e^x+1} + C$. Option B is the negative of the correct answer, a sign error. Option C would result from integrating $1/u$ instead of $1/u^2$. This problem tests recognition of perfect square trinomials and the Easy of substitution to rational functions.

📖 Explanation: In general, if an integral involves a composition of functions, we look for an 'inner function' whose derivative appears elsewhere in the integrand. This inner function is typically a nonlinear expression. While options B and C might be specific cases, option A is the general principle that guides the choice, and is the most comprehensive.

📖 Explanation: Let $u = x^2+3x+5$, $du = (2x+3) dx$. The integral simplifies to $\int u^{-2} du = -u^{-1} + C = -\frac{1}{x^2+3x+5} + C$. Option B is a sign error. Option C incorrectly integrates with respect to $u$ and then substitutes back. The correct approach is to recognize that the numerator is exactly the derivative of the denominator.

📖 Explanation: To find the area, evaluate the definite integral $\int_0^{\sqrt{\pi}} x \sin(x^2) dx$. Let $u = x^2$, $du = 2x dx$. When $x=0, u=0$; when $x=\sqrt{\pi}, u=\pi$. The integral becomes $\frac{1}{2} \int_0^\pi \sin u du = \frac{1}{2} [-\cos u]_0^\pi = \frac{1}{2} (1 - (-1)) = 1$. Option B is 0, which would be the result of a sign error or integration of a different function.

📖 Explanation: Let $u = x^3$, so $du = 3x^2 dx$. The integral becomes $\frac{1}{3} \int \frac{du}{\sqrt{u^2+1}} = \frac{1}{3} \sinh^{-1}(u) + C = \frac{1}{3} \sinh^{-1}(x^3) + C$. Option A is correct. Option D is an equivalent expression since $\sinh^{-1}(z) = \ln(z+\sqrt{z^2+1})$, but option A is the most direct and elegant answer.

📖 Explanation: If you set $u = x^2+1$ and incorrectly keep $dx$ instead of converting to $du$, you'd get $\int \sqrt{u} \cdot 2x dx = \int \sqrt{u} \, dx$, which is not in terms of $u$ alone. This is a common mistake, especially for students transitioning from basic to more complex substitutions. Option A clearly shows the incomplete substitution, making it the correct answer.

📖 Explanation: Let $u = x^2$, $du = 2x dx$. The limits become: when $x = \sqrt{\pi}/2$, $u = \pi/4$; when $x = \sqrt{\pi}$, $u = \pi$. The integral is $\int_{\pi/4}^{\pi} \cos u du = [\sin u]_{\pi/4}^{\pi} = \sin\pi - \sin(\pi/4) = 0 - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{2}$. Option B is the negative of the correct answer, likely from swapping limits or a sign error. This problem tests careful substitution of limits and attention to signs.

📖 Explanation: Let $u = x^2$, $du = 2x dx$. Then $\int x \sec^2(x^2) dx = \frac{1}{2} \int \sec^2 u du = \frac{1}{2} \tan u + C = \frac{1}{2} \tan(x^2) + C$. Option B is missing the $1/2$ factor. Option C incorrectly integrates a derivative of a different trigonometric function. Option D has the same issue with the factor of 1/2 and uses the wrong trigonometric function.

📖 Explanation: Let $u = 4-x^2$, $du = -2x dx$. Then $x dx = -\frac{1}{2} du$. The integral becomes $-\frac{1}{2} \int u^{3/2} du = -\frac{1}{5} u^{5/2} + C = -\frac{1}{5}(4-x^2)^{5/2} + C$. Option A is correct. Option B would be the result if the negative sign was forgotten. Option C or D would be the result of a factor error.

📖 Explanation: The fundamental theorem of calculus states that differentiation and integration are inverse processes. The best way to verify an antiderivative is to differentiate it and see if you get back the original integrand. Differentiating $\ln|\ln x|$ gives $\frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x\ln x}$, confirming the answer. Option B suggests a different approach but isn't a verification method. Option C is unrelated.

📖 Explanation: The integrand's structure $x^3 dx$ is almost a multiple of the derivative of $x^4$, but the presence of $x^4$ under the square root suggests $u = 1-x^4$. Then $du = -4x^3 dx$, exactly matching the $x^3 dx$ part. Option A would make $du = 4x^3 dx$, which would require adjusting signs. Option B is the most direct and cleaner substitution. Option C would involve more complex differentiation.

📖 Explanation: The integral involves $x e^{x^2}$. The derivative of $x^2$ is $2x dx$, which is present. Setting $u = 2x^2$ gives $du = 4x dx$, which also works. However, for $x^2 e^{2x^2}$, letting $u = 2x^2$ gives $du = 4x dx$ and the integrand becomes $x^2 e^u = (u/2) e^u$, making the integral straightforward. Option A leaves a $u/2$ as well, but B is more direct for the exponential argument. Option C would require solving for $dx$ in terms of $u$ and $x$, making it messy.

📖 Explanation: The solution shows a classic error of not completely substituting all parts of the integrand. If $u = x^2$, then $du = 2x dx$, so $x dx = \frac{1}{2} du$. The correct integral is $\frac{1}{2} \int \cos u du = \frac{1}{2} \sin(x^2) + C$. The student omitted the factor of $1/2$. Option A accurately identifies the missing factor. Option B is false because the substitution is appropriate.

📖 Explanation: Let $u = \sin x$, $du = \cos x dx$. When $x=0, u=0$; when $x=\pi/2, u=1$. The integral becomes $\int_0^1 u^3 du = \frac{u^4}{4} \big|_0^1 = \frac{1}{4}$. Option B or D would be the result of integrating a different power. Option C would be the result of a sign error. This problem tests substitution with definite limits and a simple polynomial after substitution.

📖 Explanation: The integral has $x^2 dx$ and a square root involving $(x^3)^2$. Let $u = x^3$, then $du = 3x^2 dx$. The integral becomes $\frac{1}{3} \int \frac{du}{\sqrt{1-u^2}} = \frac{1}{3} \sin^{-1}(u) + C = \frac{1}{3} \sin^{-1}(x^3) + C$. Option B correctly identifies the inner function that, when squared, matches the radical term. Option A would not simplify the radical. Option C would require solving for $x$ and $dx$, which is cumbersome.

📖 Explanation: While numerical methods can approximate the area, $u$-substitution provides an exact analytical solution for this function. Since the derivative of $x^2$ is $2x$, the substitution is straightforward. Relying solely on graph reading or Riemann sums would be inefficient and less accurate for such a function with a clear substitution pattern. Option A is the most mathematically sound approach.

📖 Explanation: Let $u = \cos x$, $du = -\sin x dx$. The integral becomes $-\int u^{-3} du = \frac{1}{2} u^{-2} + C = \frac{1}{2} \sec^2 x + C$. Option A is correct. Option B is an alternative form since \frac{1}{2} \tan^2 x + C = \frac{1}{2}(\sec^2 x - 1) + C = \frac{1}{2} \sec^2 x + C', but in terms of $u$, option A is the most direct. Option C has a sign error. This problem tests substitution with trigonometric functions and power reduction.