📝 Trig substitution for quadratic expressions (38 MCQs)

📖 Explanation: The denominator $x^2-6x+13$ is an irreducible quadratic ($b^2-4ac = 36-52 = -16 < 0$). Therefore, factoring is impossible, and completing the square is the standard first step to rewrite it as $(x-3)^2+4$, enabling a trigonometric or arctangent substitution. Long division applies only to improper rational functions.

📖 Explanation: Completing the square: $x^2-4x+5 = (x-2)^2+1$. Then $\int \frac{dx}{(x-2)^2+1} = \tan^{-1}(x-2)+C$. The coefficient of the squared term in the completed square form is 1, so the arctangent formula applies directly without an extra factor. Option A would be correct for $\int \frac{dx}{(x-2)^2+4}$.

📖 Explanation: First complete the square: $x^2-4x+8 = (x-2)^2+4$. Let $u=x-2$, so $x=u+2$, $dx=du$. The numerator $2x-3 = 2u+1$. The integral becomes $\int \frac{2u+1}{u^2+4} du = \ln(u^2+4) + \frac{1}{2}\tan^{-1}(u/2)+C$. Back substituting gives option A. Note the logarithmic term has coefficient 1/2 from the substitution $du = dx$.

📖 Explanation: The completed square form $(x+1)^2+4$ is of the type $u^2 + a^2$ with $u=x+1$ and $a=2$. The standard trigonometric substitution for $\sqrt{u^2+a^2}$ is $u = a\tan\theta$, which simplifies $\sqrt{u^2+a^2}$ to $a\sec\theta$. Option A $u=2\sin\theta$ is used for $\sqrt{a^2-u^2}$, and Option C lacks the constant $a$ needed for proper simplification.

📖 Explanation: The formula \int \frac{f&#039;(x)}{f(x)} dx = \ln|f(x)|+C applies only when the numerator is exactly the derivative of the denominator. Here the derivative of $x^2+2x+2$ is $2x+2$, not 1. The denominator needs to be completed to a square $(x+1)^2+1$, leading to an arctangent result: $\tan^{-1}(x+1)+C$, not a logarithm.

📖 Explanation: Complete the square: $x^2-6x+18 = (x-3)^2+9$. Using the formula $\int \frac{dx}{u^2+a^2} = \frac{1}{a}\tan^{-1}(u/a)+C$ with $u=x-3$, $a=3$, the result is $\frac{1}{3}\tan^{-1}\left(\frac{x-3}{3}\right)+C$. The coefficient $1/3$ is essential; omitting it (option C) is a common error.

📖 Explanation: The numerator $x+1$ is exactly one-half the derivative of $x^2+2x+10$ (derivative is $2x+2 = 2(x+1)$). Thus, the substitution $u=x^2+2x+10$ simplifies the integral to $\frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u|+C$. This is more efficient than completing the square and using a trigonometric substitution, demonstrating the importance of recognizing the derivative relationship.

📖 Explanation: For a proper rational function with distinct linear and irreducible quadratic factors, the decomposition assigns a constant numerator $A$ to the linear factor and a linear numerator $Bx+C$ to the quadratic factor. The quadratic $x^2+2x+2$ has discriminant $4-8 = -4 < 0$, so it is irreducible. Option B is wrong because the quadratic partial fraction must have a numerator of degree one less than the denominator ($Bx+C$).

📖 Explanation: The denominator $x^2+4x+5$ has discriminant $16-20 = -4 < 0$, so it is an irreducible quadratic. After completing the square $(x+2)^2+1$, the integral becomes $\tan^{-1}(x+2)+C$. Partial fractions only apply when the denominator can be factored into linear factors over the reals (discriminant ≥ 0), and this is not an improper or trigonometric integral in the sense of $u=\tan(x/2)$ substitution.

📖 Explanation: Complete the square: $x^2+2x+a = (x+1)^2 + (a-1)$. For convergence, the denominator must have no real roots, so $a-1 > 0 \Rightarrow a>1$. Then $\int_{-\infty}^{\infty} \frac{dx}{(x+1)^2 + (a-1)} = \frac{\pi}{\sqrt{a-1}}$. Setting this equal to $\pi$ gives $\frac{1}{\sqrt{a-1}} = 1 \Rightarrow a=2$. Option C $a=0$ gives a divergent integral (pole at $x=0, -2$), and Option A $a=1$ gives a pole at $x=-1$.

📖 Explanation: Completing the square for $x^2-2x$: add and subtract $(\frac{-2}{2})^2 = 1$: $x^2-2x = (x^2-2x+1)-1 = (x-1)^2-1$. This is of the form $u^2-a^2$ with $u=x-1$, $a=1$, suggesting a secant substitution $u = a\sec\theta$. Option A is incorrect because the sign is wrong; it would be $+1$ if the quadratic were $x^2-2x+2$.

📖 Explanation: The denominator $x^2+2x+5$ is irreducible (discriminant $4-20=-16<0$). Thus, it cannot be factored into distinct linear factors over the reals, making partial fractions inapplicable. The standard method is to complete the square: $(x+1)^2+4$, leading to $\frac{1}{2}\tan^{-1}(\frac{x+1}{2})+C$. A $u$-substitution alone would not work because the numerator is 1, not $2x+2$.

📖 Explanation: Complete the square: $x^2+2x+2 = (x+1)^2+1$. Then $\int \frac{dx}{(x+1)^2+1} = \tan^{-1}(x+1)+C$. The denominator in the completed square form is $u^2+1$ with $u=x+1$, so the coefficient $1/a$ in $\frac{1}{a}\tan^{-1}(u/a)$ is 1. Option C is incorrect because the numerator is not the derivative of the denominator. Option D uses $x$ instead of $x+1$, a common shift error.

📖 Explanation: The denominator $x^2+4x+3$ has discriminant $16-12=4 \ge 0$, so it factors into $(x+1)(x+3)$. Since the integrand is a proper rational function, the preferred first step is to factor the denominator and then apply partial fractions. Completing the square is less efficient in this case, and long division is not needed for a proper fraction. The substitution $u=x^2+4x+3$ would not simplify because the numerator is not the derivative.

📖 Explanation: Factor $x^2-4x+3 = (x-1)(x-3)$. For distinct linear factors, each contributes a term with a constant numerator: $\frac{A}{x-1} + \frac{B}{x-3}$. Option B incorrectly factors as $(x+1)(x-3)$, which gives $x^2-2x-3$. Option C is also an incorrect factorization. Option D is the form for an irreducible quadratic, but this quadratic is reducible (discriminant $16-12=4>0$).

📖 Explanation: The numerator $2x-2$ is exactly the derivative of $x^2-2x+5$. Thus, $\int \frac{2x-2}{x^2-2x+5} dx = \ln|x^2-2x+5|+C$. Since $x^2-2x+5 = (x-1)^2+4 > 0$ for all real $x$, absolute value bars are unnecessary but not incorrect. Option B is the result if the numerator were 1. Option C has an incorrect coefficient of 1/2. Option D incorrectly adds an arctangent term.

📖 Explanation: $x^2+6x+10 = (x^2+6x+9)+1 = (x+3)^2+1$. This is exactly the form $u^2+1$ with $u=x+3$, leading to $\tan^{-1}(x+3)+C$. Option B uses the wrong sign (should be +1). Option C has the wrong center of the square; it should be $x+3$, not $x-3$.

📖 Explanation: The denominator $x^2+4x+4 = (x+2)^2$ has a repeated linear factor. The integral becomes $\int \frac{dx}{(x+2)^2} = -\frac{1}{x+2}+C$. It is a proper rational function (degree of numerator 0 < degree of denominator 2). Option B is false because the discriminant is 0, not negative. Option C is false because the interval of integration is not specified as improper, and the integral converges on any finite interval not containing $x=-2$.

📖 Explanation: The substitution $u=x^2+2x+2$ gives $du=(2x+2)dx$, but the numerator is $x$, not $2x+2$. The correct approach is to complete the square: $x^2+2x+2 = (x+1)^2+1$, then write $x = (x+1)-1$. The integral splits into $\int \frac{x+1}{(x+1)^2+1} dx - \int \frac{1}{(x+1)^2+1} dx = \frac{1}{2}\ln((x+1)^2+1) - \tan^{-1}(x+1)+C$.

📖 Explanation: The derivative of $x^2-2x+3$ is $2x-2 = 2(x-1)$. The numerator $x-1$ is exactly half of the derivative. Thus, $\int \frac{x-1}{x^2-2x+3} dx = \frac{1}{2}\int \frac{2x-2}{x^2-2x+3} dx = \frac{1}{2}\ln(x^2-2x+3)+C$. Option C incorrectly adds an arctangent term; this only appears if the numerator is not proportional to the derivative. Option D has the wrong coefficient and misses the 1/2 factor.

📖 Explanation: The sign of the discriminant determines the factorization of the quadratic. If $b^2-4ac > 0$, it factors into distinct linear factors, so partial fractions is the natural method. If $=0$, it's a perfect square (repeated linear factor), and the integral is straightforward. If $<0$, it's irreducible, so completing the square and using an arctangent substitution is appropriate. Options B and C are incorrect. This Medium framework is central to choosing the integration method.

📖 Explanation: Complete the square: $x^2-2x+2 = (x-1)^2+1$. Then $\int_0^1 \frac{dx}{(x-1)^2+1} = [\tan^{-1}(x-1)]_0^1 = \tan^{-1}(0) - \tan^{-1}(-1) = 0 - (-\frac{\pi}{4}) = \frac{\pi}{4}$. Option A has the wrong value, calculating the integral from -1 to 1. Option C uses the wrong order of subtraction. Option D uses an incorrect antiderivative; the logarithm form would require the numerator to be $2x-2$.

📖 Explanation: The integrand in B has an irreducible quadratic in the denominator with a linear term. Completing the square transforms it into $(x+2)^2+1$, making it integrable via arctangent. Integral A already matches the form $u^2+a^2$ without completing the square. Integral C is best handled by partial fractions because the denominator factors. Integral D already has a simple denominator with a repeated linear factor, best handled by substitution or partial fractions.

📖 Explanation: The derivative of $x^2+6x+10$ is $2x+6 = 2(x+3)$. The numerator $x+3$ is exactly half of this derivative. Thus, $\int \frac{x+3}{(x+3)^2+1} dx = \frac{1}{2}\int \frac{2x+6}{(x+3)^2+1} dx = \frac{1}{2}\ln((x+3)^2+1)+C$. There is no need to split further. The choice of $u=x^2+6x+10$ works directly. Option B, C, or D would be incorrect because they introduce extra terms.

📖 Explanation: $-x^2+4x = -(x^2-4x) = -(x^2-4x+4-4) = -((x-2)^2-4) = 4-(x-2)^2$. This is of the form $a^2-u^2$ with $a=2$, $u=x-2$, suggesting a sine substitution. This type arises from quadratic expressions with a negative leading coefficient. Option B has the wrong sign, and Options C and D would result from $x^2+4x+$ constant.

📖 Explanation: Complete the square: $-x^2+6x = 9-(x-3)^2$. Then $\int \frac{dx}{\sqrt{9-(x-3)^2}} = \sin^{-1}\left(\frac{x-3}{3}\right)+C$. The formula $\int \frac{du}{\sqrt{a^2-u^2}} = \sin^{-1}(u/a)+C$ applies. Option B is the result for $\int \frac{du}{\sqrt{u^2-a^2}}$. Option C is missing the factor $1/3$ inside the arcsine. Option D is for $\int \frac{du}{a^2+u^2}$.

📖 Explanation: Complete the square: $x^2-4x+8 = (x-2)^2+4$. The area is $\int_0^2 \frac{dx}{(x-2)^2+4} = \left[\frac{1}{2}\tan^{-1}\left(\frac{x-2}{2}\right)\right]_0^2$. Evaluating: at x=2, $\frac{1}{2}\tan^{-1}(0) = 0$. At x=0, $\frac{1}{2}\tan^{-1}(-1) = -\frac{\pi}{8}$. So the area is $0 - (-\frac{\pi}{8}) = \frac{\pi}{8}$. Option A calculates the integral from -2 to 2. This question models finding area under a rational curve involving an irreducible quadratic.

📖 Explanation: The student attempted to use a formula for $\int \frac{dx}{x^2-a^2}$, which yields a logarithm. However, $x^2-4x+5 = (x-2)^2+1$, which is of the form $u^2+a^2$, not $u^2-a^2$. The correct antiderivative is $\tan^{-1}(x-2)+C$. The discriminant $16-20=-4<0$ confirms it is irreducible and should not have been factored. This is a classic error of misidentifying the type of quadratic.

📖 Explanation: Write numerator as $3x+4 = \frac{3}{2}(2x+4) - 2$. Then $\int \frac{3x+4}{x^2+4x+5} dx = \frac{3}{2}\int \frac{2x+4}{x^2+4x+5} dx - 2\int \frac{dx}{(x+2)^2+1}$. The first integral is $\frac{3}{2}\ln(x^2+4x+5) - 2\tan^{-1}(x+2)+C$. Option A matches this. Option B has the wrong sign for the arctangent term. Option C has the wrong coefficient for the log. Option D has the wrong coefficient for the arctangent.

📖 Explanation: Complete the square: $x^2-2x+2 = (x-1)^2+1$, so $y = \frac{1}{(x-1)^2+1}$. This function is maximum when the denominator is minimum. The denominator $(x-1)^2+1$ is minimized at $x=1$, with value 1. Thus, the maximum is $y=1$. This corresponds to point (1,1). Option B gives the value at the inflection point of the denominator's derivative. Option C is incorrect. Option D is false because the function is bounded above by 1.

📖 Explanation: The integral $\int \frac{dx}{(x+1)^2+1}$ can be evaluated by $u=x+1$ giving $\tan^{-1}u+C$. It can also be evaluated by $u=x^2+2x+2$ if the numerator were $2x+2$, but here the numerator is 1, so this would not work. However, the substitution $x+1 = \tan\theta$ is a trigonometric substitution that also works because $dx = \sec^2\theta d\theta$, and $\sqrt{(x+1)^2+1} = \sec\theta$. But the problem asks for substitutions that correctly evaluate the integral; only A and C are correct. Option D is incorrect because substitution B does not simplify the integral.

📖 Explanation: Complete the square: $(x+1)^2+1$. Let $x+1 = \tan\theta$, $dx = \sec^2\theta d\theta$, $x^2+2x+2 = \sec^2\theta$. The integral becomes $\int \frac{\sec^2\theta}{\sec^4\theta} d\theta = \int \cos^2\theta d\theta = \frac{1}{2}(\theta + \sin\theta\cos\theta)$. Back substituting: $\theta = \tan^{-1}(x+1)$, $\sin\theta = \frac{x+1}{\sqrt{x^2+2x+2}}$, $\cos\theta = \frac{1}{\sqrt{x^2+2x+2}}$. Thus, $\sin\theta\cos\theta = \frac{x+1}{x^2+2x+2}$. Option A matches. Option C has the wrong sign, and Option B lacks the $x+1$ in the numerator.

📖 Explanation: The amount added is $\int_0^4 \frac{100}{t^2-4t+20} dt$. Complete the square: $t^2-4t+20 = (t-2)^2+16$. So the integral is $100 \int_0^4 \frac{dt}{(t-2)^2+16} = 100 \left[\frac{1}{4}\tan^{-1}\left(\frac{t-2}{4}\right)\right]_0^4 = 25\left[\tan^{-1}\left(\frac{t-2}{4}\right)\right]_0^4 = 25\left[\tan^{-1}(0.5) - \tan^{-1}(-0.5)\right] = 25(2\tan^{-1}(0.5))$. This is equal to $25\pi$ only if $\tan^{-1}(0.5) = \pi/8$, which is false. The correct numerical value is $25(2\tan^{-1}(0.5)) \approx 25(0.9273) = 23.18$. None of the options are exactly correct, but Option C gives $25\pi \approx 78.5$, which is wrong. This question tests applying the integral to a rate problem and evaluating the arctangent.

📖 Explanation: Since $\tan^{-1}(z) + \cot^{-1}(z) = \frac{\pi}{2}$, we have $\cot^{-1}(z) = \frac{\pi}{2} - \tan^{-1}(z)$. Thus, $-\cot^{-1}(z) = \tan^{-1}(z) - \frac{\pi}{2}$. The constant $\frac{\pi}{2}$ can be absorbed into the arbitrary constant of integration $C$. Therefore, both expressions represent the same family of antiderivatives. Student B's expression is valid and equivalent to Student A's. Option C is correct. This highlights the role of the constant of integration in indefinite integrals.

📖 Explanation: Using the disk method, the volume is $V = \pi \int_0^1 [f(x)]^2 dx = \pi \int_0^1 \frac{1}{(x^2+2x+5)^2} dx$. The integrand involves the square of a quadratic, which may require further techniques like completing the square and trigonometric substitution. Option B is the area under the curve, not the volume. Option C is missing the $\pi$ factor. Option D is the shell method formula for rotation about the y-axis.

📖 Explanation: Let $u=x-1$, then $du=dx$, and $x^2-2x+10 = (x-1)^2+9 = u^2+9$. The integral becomes $\int \frac{du}{u^2+9} = \frac{1}{3}\tan^{-1}\left(\frac{u}{3}\right)+C = \frac{1}{3}\tan^{-1}\left(\frac{x-1}{3}\right)+C$. Option B is the antiderivative in terms of $u$, but the problem asks for the final answer in terms of $x$. Option C misses the factor $1/3$. Option D uses the wrong formula; this is an arctangent, not a logarithm.