Definite and Indefinite Integrals Questions and Answers

Answer: B

The integral is \(\frac{1}{2}[\log(1 + x)]^{2} + C\).

Substitute \(t = \log(1 + x)\), so \(dt = \frac{dx}{1 + x}\):

\[ \begin{aligned} \int t\,dt &= \frac{t^{2}}{2} + C \\ &= \frac{1}{2}[\log(1 + x)]^{2} + C \end{aligned} \]

Without the \(\frac{1}{2}\) the derivative would be twice the integrand. \(\frac{1}{2}\log(1 + x)\) and \(\log(1 + x)\) are the integrals of \(\frac{1}{1 + x}\)-type functions, not of this one.

The general pattern is \(\int f(x)f'(x)\,dx = \frac{[f(x)]^{2}}{2} + C\), here with \(f(x) = \log(1 + x)\).

Answer: E

The integral is \(\frac{1}{2\sqrt{2}}\log\Big|\frac{\sqrt{2}+\sin\theta}{\sqrt{2}-\sin\theta}\Big|\) + C.

Put \(t = \sin\theta\), \(dt = \cos\theta\,d\theta\), and use \(\int \frac{dt}{a^{2} - t^{2}} = \frac{1}{2a}\log\left|\frac{a + t}{a - t}\right|\) with \(a = \sqrt{2}\):

\[ \begin{aligned} &\int \frac{dt}{2 - t^{2}} \\ &= \frac{1}{2\sqrt{2}}\log\left|\frac{\sqrt{2} + t}{\sqrt{2} - t}\right| + C \end{aligned} \]

The options with \(\frac{1}{2}\), 1 or \(\frac{1}{\sqrt{2}}\) in front use the wrong constant \(\frac{1}{2a}\), and option A also inverts the fraction, which changes the sign.

Answer: D

The integral is \(\frac{\pi x}{2} + C\).

For \(0 \le \sqrt{x} \le 1\), \(\sin^{-1}\sqrt{x} + \cos^{-1}\sqrt{x} = \frac{\pi}{2}\), a constant. So

\[ \int \frac{\pi}{2}\,dx = \frac{\pi x}{2} + C \]

\(\frac{\pi}{2} + C\) forgets to integrate the constant, and the other multiples of \(\pi x\) use the wrong value of the sum of the inverse functions.

Answer: A

The integral is \(\frac{e^{x}}{1 + x} + C\).

The integrand has the form \(e^{x}[f(x) + f'(x)]\) with \(f(x) = \frac{1}{1 + x}\), since \(f'(x) = -\frac{1}{(1 + x)^{2}}\). And

\[ \int e^{x}[f(x) + f'(x)]\,dx = e^{x}f(x) + C \]

because \(\frac{d}{dx}[e^{x}f(x)] = e^{x}(f + f')\). The other options do not differentiate back to the integrand; for example \(\frac{xe^{x}}{1 + x}\) has derivative \(\frac{e^{x}(x^{2} + x + 1)}{(1 + x)^{2}}\).

The pattern \(e^{x}[f(x) + f'(x)]\) is worth spotting quickly: whenever the bracket contains a function and its derivative, the integral is \(e^{x}\) times that function.

Answer: A

The integral is \(\log\left(\frac{10}{9}\right)\).

Use partial fractions \(\frac{1}{x(1 + x)} = \frac{1}{x} - \frac{1}{1 + x}\):

\[ \begin{aligned} \Big[\log\frac{x}{1 + x}\Big]_{3}^{5} &= \log\frac{5}{6} - \log\frac{3}{4} \\ &= \log\frac{20}{18} = \log\frac{10}{9} \end{aligned} \]

\(\log 5\) and \(\log 2\) integrate only one partial fraction, and \(\frac{11}{9}\) and \(\frac{13}{9}\) come from evaluating at the wrong limits.

A quick check: the integrand is about \(\frac{1}{16}\) at the middle of [3, 5], so the integral is about \(2 \times \frac{1}{16} \approx 0.12\), close to \(\log\frac{10}{9} \approx 0.105\) and far below \(\log 2\) or \(\log 5\).

Answer: A

The integral is −3.

\([x]\) is constant on each unit interval, so split the range:

\[ \begin{aligned} \int_{-3}^{3}[x]\,dx &= (-3) + (-2) + (-1) + 0 + 1 + 2 \\ &= -3 \end{aligned} \]

The answer is not 0 because \([x]\) is not an odd function: on [−3, −2) it is −3, but on [2, 3) it is only 2. −6 and −4 miscount the intervals.

Answer: E

The integral is 0.

\(x^{5} + x^{3} + x\) is odd and \(\cos x\) is even, so their product is odd: \(g(-x) = -g(x)\). The integral of an odd function over a symmetric interval \([-a, a]\) is 0.

The non-zero options would require actually integrating, but the contributions from \([-\frac{\pi}{2}, 0]\) and \([0, \frac{\pi}{2}]\) cancel exactly.

Answer: C

The integral is 4.

\(e^{|x|}\) is even, so

\[ \begin{aligned} \int_{-\log 3}^{\log 3} e^{|x|}dx &= 2\int_{0}^{\log 3} e^{x}dx \\ &= 2(e^{\log 3} - 1) = 2(3 - 1) = 4 \end{aligned} \]

3 is \(e^{\log 3}\) alone, 6 forgets to subtract \(e^{0} = 1\), and 0 wrongly treats the integrand as odd.

Answer: D

The integral is \(-3\tan^{-1/3}x + C\).

The powers add to 2, so divide numerator and denominator by \(\cos^{2}x\):

\[ \begin{aligned} \frac{1}{\cos^{2/3}x\,\sin^{4/3}x} &= \frac{\sec^{2}x}{\tan^{4/3}x} \\ \int t^{-4/3}\,dt &= -3t^{-1/3} + C \end{aligned} \]

with \(t = \tan x\), \(dt = \sec^{2}x\,dx\). So the answer is \(-3\tan^{-1/3}x + C\). Here \(\tan^{-1/3}\) means a power, not an inverse function. \(3\tan^{-1/3}x\) misses the minus sign from integrating a negative power, and the options with \(\tan^{1/3}x\) have the wrong exponent.

Answer: A

The integral is \(\frac{(1 - x)^{12}}{12} - \frac{(1 - x)^{11}}{11} + C\).

Substitute \(u = 1 - x\), so \(x = 1 - u\) and \(dx = -du\):

\[ \begin{aligned} \int x(1 - x)^{10}dx &= -\int (1 - u)u^{10}du \\ &= \int (u^{11} - u^{10})du \\ &= \frac{u^{12}}{12} - \frac{u^{11}}{11} + C \end{aligned} \]

Putting back \(u = 1 - x\) gives the answer. The '+' version comes from losing the minus sign of dx, and the options with \((1 + x)^{11}\) use the wrong substitution.