Use integration by parts:
Let \( u = \ln(2x) \), \( dv = x \, dx \)
Then \( du = \frac{1}{x} \cdot 2 \cdot \frac{1}{2} = \frac{1}{x} \, dx \), and \( v = \frac{x^2}{2} \)

Apply the integration by parts formula: \[ \int x \ln(2x) \, dx = \frac{x^2}{2} \ln(2x) - \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx \] Simplify the remaining integral: \[ \int \frac{x^2}{2} \cdot \frac{1}{x} \, dx = \int \frac{x}{2} \, dx = \frac{1}{2} \cdot \frac{x^2}{2} = \frac{x^2}{4} \] Final result: \[ \int x \ln(2x) \, dx = \frac{x^2}{2} \ln(2x) - \frac{x^2}{4} + C \] \[ \boxed{ \frac{x^2}{2} \ln(2x) - \frac{x^2}{4} + C } \]

Use trigonometric substitution:
Let \( x = 3\sin\theta \), so \( dx = 3\cos\theta\, d\theta \), and \( \sqrt{9 - x^2} = \sqrt{9 - 9\sin^2\theta} = 3\cos\theta \)

Substitute: \[ \int \frac{x^2}{\sqrt{9 - x^2}} \, dx = \int \frac{(3\sin\theta)^2}{3\cos\theta} \cdot 3\cos\theta\, d\theta = \int 9\sin^2\theta \, d\theta \] Use identity \( \sin^2\theta = \frac{1 - \cos(2\theta)}{2} \): \[ \int 9\sin^2\theta \, d\theta = 9 \int \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{9}{2} \int (1 - \cos(2\theta)) \, d\theta \] \[ = \frac{9}{2} \left( \theta - \frac{1}{2} \sin(2\theta) \right) + C \] Return to x using \( \theta = \sin^{-1}\left(\frac{x}{3}\right) \) and \( \sin(2\theta) = 2\sin\theta\cos\theta = 2 \cdot \frac{x}{3} \cdot \frac{\sqrt{9 - x^2}}{3} = \frac{2x\sqrt{9 - x^2}}{9} \)

Final answer: \[ \boxed{ \frac{9}{2} \sin^{-1}\left(\frac{x}{3}\right) - \frac{x\sqrt{9 - x^2}}{2} + C } \]

First, simplify using polynomial division: \[ \frac{x^2 + x + 1}{x + 1} = x + \frac{1}{x + 1} \] So the integral becomes: \[ \int_0^1 \left( x + \frac{1}{x + 1} \right) dx = \int_0^1 x \, dx + \int_0^1 \frac{1}{x + 1} \, dx \] Evaluate each part: \[ \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1}{2} \] \[ \int_0^1 \frac{1}{x + 1} \, dx = \left[ \ln|x + 1| \right]_0^1 = \ln(2) - \ln(1) = \ln 2 \] Combine: \[ \boxed{ \frac{1}{2} + \ln 2 } \]

Factor the denominator: \[ x^2 + x - 6 = (x + 3)(x - 2) \] Use partial fractions: \[ \frac{4x + 5}{(x + 3)(x - 2)} = \frac{A}{x + 3} + \frac{B}{x - 2} \] Multiply both sides by \( (x + 3)(x - 2) \): \[ 4x + 5 = A(x - 2) + B(x + 3) \] Expand: \[ 4x + 5 = Ax - 2A + Bx + 3B = (A + B)x + (-2A + 3B) \] Match coefficients: \[ A + B = 4,\quad -2A + 3B = 5 \] Solve the system:
From \( A + B = 4 \Rightarrow A = 4 - B \)
Substitute into second equation: \[ -2(4 - B) + 3B = 5 \Rightarrow -8 + 2B + 3B = 5 \Rightarrow 5B = 13 \Rightarrow B = \frac{13}{5},\quad A = \frac{7}{5} \] Now integrate: \[ \int \frac{7/5}{x + 3} + \frac{13/5}{x - 2} \, dx = \frac{7}{5} \ln|x + 3| + \frac{13}{5} \ln|x - 2| + C \] Final answer: \[ \boxed{ \frac{7}{5} \ln|x + 3| + \frac{13}{5} \ln|x - 2| + C } \]

First simplify the integrand using partial fractions: \[ \frac{3}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \] Multiply both sides by \( x(x + 1) \): \[ 3 = A(x + 1) + Bx = Ax + A + Bx = (A + B)x + A \] Match coefficients: \[ A + B = 0,\quad A = 3 \Rightarrow B = -3 \] So: \[ \int_1^{\infty} \frac{3}{x(x + 1)} \, dx = \int_1^{\infty} \left( \frac{3}{x} - \frac{3}{x + 1} \right) dx \] Now take the limit: \[ \lim_{t \to \infty} \left[ 3\ln|x| - 3\ln|x + 1| \right]_1^t = \lim_{t \to \infty} \left( 3\ln t - 3\ln(t + 1) - 3\ln 1 + 3\ln 2 \right) \] Use: \[ \ln t - \ln(t + 1) = \ln\left(\frac{t}{t + 1}\right) \Rightarrow \ln\left(1 - \frac{1}{t + 1}\right) \to 0 \text{ as } t \to \infty \] So: \[ \lim_{t \to \infty} 3 \ln\left(\frac{t}{t + 1}\right) + 3 \ln 2 = 0 + 3 \ln 2 \] Final answer: \[ \boxed{3 \ln 2} \]

Use the disk method:
The radius of each disk is \( y = \sqrt{x} \), so: \[ V = \pi \int_0^4 (\sqrt{x})^2 \, dx = \pi \int_0^4 x \, dx \] \[ \int_0^4 x \, dx = \left[ \frac{x^2}{2} \right]_0^4 = \frac{16}{2} = 8 \] Multiply by \( \pi \): \[ \boxed{8\pi} \]

Use the shell method:
Shell radius = \( x \), shell height = \( x - x^2 \)
Find limits of intersection: \( x = x^2 \Rightarrow x(x - 1) = 0 \Rightarrow x = 0, 1 \)
Set up the integral: \[ V = 2\pi \int_0^1 x(x - x^2) \, dx = 2\pi \int_0^1 (x^2 - x^3) \, dx \] \[ \int_0^1 x^2 - x^3 \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \] Multiply by \( 2\pi \): \[ \boxed{ \frac{\pi}{6} } \]

This is the alternating harmonic series.
The terms \( \frac{1}{n} \) decrease and go to zero, so by the Alternating Series Test, it converges.
However, the harmonic series \( \sum \frac{1}{n} \) diverges, so it does not converge absolutely. \[ \boxed{\text{Converges conditionally}} \]

Use the integral test: \[ \int_2^\infty \frac{1}{x(\ln x)^2} dx \] Let \( u = \ln x \Rightarrow du = \frac{1}{x} dx \) \[ \int_2^\infty \frac{1}{x(\ln x)^2} dx = \int_{\ln 2}^\infty \frac{1}{u^2} du = \left[ -\frac{1}{u} \right]_{\ln 2}^\infty = \frac{1}{\ln 2} \] Converges. So does the series: \[ \boxed{\text{Converges}} \]

Use ratio test: \[ \left| \frac{(x - 2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x - 2)^n} \right| = \left| \frac{(x - 2) \cdot n}{3(n+1)} \right| \to \frac{|x - 2|}{3} \] Converges if: \[ \frac{|x - 2|}{3} < 1 \Rightarrow |x - 2| < 3 \Rightarrow -1 < x < 5 \] Check endpoints: At \( x = -1 \): \( \sum \frac{(-3)^n}{n \cdot 3^n} = \sum \frac{(-1)^n}{n} \Rightarrow \) converges conditionally At \( x = 5 \): \( \sum \frac{3^n}{n \cdot 3^n} = \sum \frac{1}{n} \Rightarrow \) diverges Final interval: \[ \boxed{(-1, 5)} \]

Known expansion: \[ \ln(1 + x) = \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n} \] Radius of convergence: \( R = 1 \Rightarrow \) interval: \( (-1, 1] \) Final: \[ \boxed{\sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n},\quad x \in (-1, 1]} \]

Separate variables: \[ \frac{dy}{y^2} = x dx \] Integrate both sides: \[ \int y^{-2} dy = \int x dx \Rightarrow -\frac{1}{y} = \frac{x^2}{2} + C \] Solve for \( y \): \[ y = -\frac{1}{\frac{x^2}{2} + C} \] Use initial condition \( y(0) = 1 \): \[ 1 = -\frac{1}{0 + C} \Rightarrow C = -1 \] Final: \[ \boxed{y(x) = -\frac{1}{\frac{x^2}{2} - 1}} \]

Arc length: \[ L = \int_0^2 \sqrt{(dx/dt)^2 + (dy/dt)^2} dt \] \[ dx/dt = 2,\quad dy/dt = 6t \Rightarrow L = \int_0^2 \sqrt{2^2 + (6t)^2} dt = \int_0^2 \sqrt{4 + 36t^2} dt \] Use substitution: \( u = 6t \Rightarrow du = 6 dt \Rightarrow dt = \frac{du}{6} \) \[ t = 0 \Rightarrow u = 0,\quad t = 2 \Rightarrow u = 12 \] \[ L = \int_0^{12} \sqrt{4 + u^2} \cdot \frac{1}{6} du = \frac{1}{6} \int_0^{12} \sqrt{4 + u^2} du \] Use trig substitution or numerical: \[ \int \sqrt{4 + u^2} du = \frac{u}{2} \sqrt{4 + u^2} + 2 \ln(u + \sqrt{4 + u^2}) \] Evaluate at 12 and 0: \[ \Rightarrow \boxed{L \approx 25.15} \]

Loop goes from \( \theta = 0 \) to \( \theta = \pi \) \[ A = \frac{1}{2} \int_0^\pi (1 + \cos \theta)^2 d\theta \] Expand: \[ (1 + \cos \theta)^2 = 1 + 2\cos \theta + \cos^2 \theta \] Use identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] Integrate: \[ \frac{1}{2} \int_0^\pi \left(1 + 2\cos \theta + \frac{1 + \cos 2\theta}{2} \right) d\theta \] Simplify: \[ \frac{1}{2} \int_0^\pi \left( \frac{3}{2} + 2\cos \theta + \frac{1}{2} \cos 2\theta \right) d\theta \] Integrate term-by-term: \[ \frac{1}{2} \left[ \frac{3}{2} \theta + 2 \sin \theta + \frac{1}{4} \sin 2\theta \right]_0^\pi = \frac{1}{2} \left[ \frac{3}{2} \pi \right] = \frac{3\pi}{4} \] \[ \boxed{ \frac{3\pi}{4} } \]

Use identities: \[ r = 2 \sin \theta,\quad y = r \sin \theta \Rightarrow y = 2 \sin^2 \theta = 2 \left( 1 - \cos^2 \theta \right) \] Also: \[ x = r \cos \theta \Rightarrow \cos \theta = \frac{x}{r} \] Better approach: \[ r = 2 \sin \theta \Rightarrow r^2 = 2r \sin \theta \Rightarrow x^2 + y^2 = 2y \Rightarrow x^2 + y^2 - 2y = 0 \Rightarrow x^2 + (y - 1)^2 = 1 \] \[ \boxed{ x^2 + (y - 1)^2 = 1 } \]

\[ dx/dt = -\sin t,\quad dy/dt = 2\cos 2t \Rightarrow \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\cos 2t}{- \sin t} \] At \( t = \frac{\pi}{4} \): \[ \cos 2t = \cos \frac{\pi}{2} = 0,\quad \sin t = \frac{\sqrt{2}}{2} \Rightarrow \frac{dy}{dx} = \frac{0}{-\frac{\sqrt{2}}{2}} = 0 \] \[ \boxed{0} \]

Separate variables: \[ (1 + y^2)dy = 2x dx \Rightarrow \int (1 + y^2) dy = \int 2x dx \] \[ y + \frac{y^3}{3} = x^2 + C \Rightarrow \boxed{ y + \frac{y^3}{3} = x^2 + C } \]

First, identify the region of integration. The curves intersect where \( x = x^2 \Rightarrow x^2 - x = 0 \Rightarrow x(x - 1) = 0 \Rightarrow x = 0, 1 \) Use the **shell method**, since we’re rotating around the y-axis: \[ V = 2\pi \int_0^1 x \cdot (x - x^2) \, dx \] \[ = 2\pi \int_0^1 (x^2 - x^3) \, dx \] Compute the integral: \[ \int_0^1 (x^2 - x^3) \, dx = \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1 = \frac{1}{3} - \frac{1}{4} = \frac{1}{12} \] Final volume: \[ V = 2\pi \cdot \frac{1}{12} = \frac{\pi}{6} \] \[ \boxed{ \frac{\pi}{6} } \]

The surface area of a curve rotated about the x-axis is: \[ S = \int_a^b 2\pi y(t) \sqrt{(dx/dt)^2 + (dy/dt)^2} \, dt \] Compute derivatives: \[ dx/dt = 1,\quad dy/dt = \frac{1}{2\sqrt{t}} \Rightarrow \left( \frac{dy}{dt} \right)^2 = \frac{1}{4t} \] Now plug into the surface area formula: \[ S = \int_1^4 2\pi \sqrt{t} \cdot \sqrt{1 + \frac{1}{4t}} \, dt \] Simplify the square root: \[ \sqrt{1 + \frac{1}{4t}} = \sqrt{ \frac{4t + 1}{4t} } = \frac{\sqrt{4t + 1}}{2\sqrt{t}} \] Multiply: \[ 2\pi \sqrt{t} \cdot \frac{\sqrt{4t + 1}}{2\sqrt{t}} = \pi \sqrt{4t + 1} \] So the integral becomes: \[ S = \pi \int_1^4 \sqrt{4t + 1} \, dt \] Use substitution: \[ u = 4t + 1 \Rightarrow du = 4 dt \Rightarrow dt = \frac{du}{4} \] When \( t = 1 \), \( u = 5 \) When \( t = 4 \), \( u = 17 \) \[ S = \pi \int_5^{17} \sqrt{u} \cdot \frac{1}{4} du = \frac{\pi}{4} \int_5^{17} u^{1/2} du \] Compute: \[ \frac{\pi}{4} \cdot \left[ \frac{2}{3} u^{3/2} \right]_5^{17} = \frac{\pi}{4} \cdot \frac{2}{3} \left( 17^{3/2} - 5^{3/2} \right) \] Approximate values: \[ 17^{3/2} = \sqrt{17}^3 \approx (4.123)^3 \approx 70.1,\quad 5^{3/2} \approx (2.236)^3 \approx 11.2 \] \[ S \approx \frac{\pi}{4} \cdot \frac{2}{3} (70.1 - 11.2) = \frac{\pi}{4} \cdot \frac{2}{3} (58.9) \approx \frac{\pi}{4} \cdot 39.27 \approx 30.84 \] Final answer: \[ \boxed{S \approx 30.84} \]