To solve \(\int_{0}^{1} xe^{2x} \, dx\), we use integration by parts. Let \(u = x\) and \(dv = e^{2x} \, dx\).

Then \(du = dx\) and \(v = \frac{1}{2}e^{2x}\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int xe^{2x} \, dx = x \cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x} \, dx\).

\(= \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} + C\).

Now, evaluate from 0 to 1:

\(\left[ \frac{1}{2}xe^{2x} - \frac{1}{4}e^{2x} \right]_{0}^{1}\).

At \(x = 1\): \(\frac{1}{2} \cdot 1 \cdot e^{2} - \frac{1}{4}e^{2} = \frac{1}{4}e^{2}\).

At \(x = 0\): \(\frac{1}{2} \cdot 0 \cdot e^{0} - \frac{1}{4}e^{0} = -\frac{1}{4}\).

Thus, the definite integral is:

\(\frac{1}{4}e^{2} - (-\frac{1}{4}) = \frac{1}{4}(e^{2} + 1)\).

To solve \(\int_{1}^{2} x \ln x \, dx\), we use integration by parts. Let \(u = \ln x\) and \(dv = x \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{1}{2} x^2\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int x \ln x \, dx = \frac{1}{2} x^2 \ln x - \int \frac{1}{2} x^2 \cdot \frac{1}{x} \, dx\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{2} \int x \, dx\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{2} \cdot \frac{1}{2} x^2 + C\)

\(= \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 + C\)

Now, evaluate from 1 to 2:

\(\left[ \frac{1}{2} x^2 \ln x - \frac{1}{4} x^2 \right]_{1}^{2}\)

\(= \left( \frac{1}{2} (2)^2 \ln 2 - \frac{1}{4} (2)^2 \right) - \left( \frac{1}{2} (1)^2 \ln 1 - \frac{1}{4} (1)^2 \right)\)

\(= \left( 2 \ln 2 - 1 \right) - \left( 0 - \frac{1}{4} \right)\)

\(= 2 \ln 2 - 1 + \frac{1}{4}\)

\(= 2 \ln 2 - \frac{3}{4}\)

Let \(u = \ln x\) and \(dv = \frac{1}{x^4} \, dx\). Then \(du = \frac{1}{x} \, dx\) and \(v = -\frac{1}{3} x^{-3}\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\).

\(\int \frac{\ln x}{x^4} \, dx = -\frac{1}{3} x^{-3} \ln x + \frac{1}{3} \int x^{-3} \cdot \frac{1}{x} \, dx\).

\(= -\frac{1}{3} x^{-3} \ln x + \frac{1}{3} \int x^{-4} \, dx\).

\(= -\frac{1}{3} x^{-3} \ln x - \frac{1}{9} x^{-3}\).

Evaluate from 1 to \(a\):

\(\left[ -\frac{1}{3} x^{-3} \ln x - \frac{1}{9} x^{-3} \right]_{1}^{a}\).

\(= \left( -\frac{1}{3} a^{-3} \ln a - \frac{1}{9} a^{-3} \right) - \left( 0 - \frac{1}{9} \right)\).

\(= \frac{1}{9} - \frac{1}{3} a^{-3} \ln a - \frac{1}{9} a^{-3}\).

Since \(a > 1\), \(a^{-3} \ln a > 0\) and \(a^{-3} > 0\), thus \(\int_{1}^{a} \frac{\ln x}{x^4} \, dx < \frac{1}{9}\).

Let \(u = \arctan\left(\frac{1}{2}x\right)\) and \(dv = dx\). Then \(du = \frac{1}{1+\left(\frac{1}{2}x\right)^2} \cdot \frac{1}{2} \, dx = \frac{1}{4+x^2} \, dx\) and \(v = x\).

Using integration by parts, \(\int u \, dv = uv - \int v \, du\), we have:

\(\int \arctan\left(\frac{1}{2}x\right) \, dx = x \arctan\left(\frac{1}{2}x\right) - \int x \cdot \frac{1}{4+x^2} \, dx\).

Now, solve \(\int x \cdot \frac{1}{4+x^2} \, dx\) using substitution. Let \(w = 4 + x^2\), then \(dw = 2x \, dx\), so \(x \, dx = \frac{1}{2} \, dw\).

Thus, \(\int x \cdot \frac{1}{4+x^2} \, dx = \frac{1}{2} \int \frac{1}{w} \, dw = \frac{1}{2} \ln |w| + C = \frac{1}{2} \ln(4+x^2) + C\).

Substitute back:

\(x \arctan\left(\frac{1}{2}x\right) - \frac{1}{2} \ln(4+x^2)\).

Evaluate from 0 to 2:

\(\left[ x \arctan\left(\frac{1}{2}x\right) - \frac{1}{2} \ln(4+x^2) \right]_0^2\).

At \(x = 2\):

\(2 \arctan(1) - \frac{1}{2} \ln(8) = 2 \cdot \frac{\pi}{4} - \frac{1}{2} \ln(8) = \frac{\pi}{2} - \ln(2)\).

At \(x = 0\):

\(0 \cdot \arctan(0) - \frac{1}{2} \ln(4) = 0 - \ln(2)\).

Thus, the exact value is:

\(\frac{\pi}{2} - \ln(2) - (0 - \ln(2)) = \frac{\pi}{2} - \ln(2)\).

To solve \(\int_{1}^{8} x^{-\frac{2}{3}} \ln x \, dx\), we use integration by parts.

Let \(u = \ln x\) and \(dv = x^{-\frac{2}{3}} \, dx\).

Then \(du = \frac{1}{x} \, dx\) and \(v = \frac{x^{\frac{1}{3}}}{\frac{1}{3}} = 3x^{\frac{1}{3}}\).

Integration by parts gives:

\(\int u \, dv = uv - \int v \, du\)

\(= 3x^{\frac{1}{3}} \ln x - 3 \int x^{\frac{1}{3}} \cdot \frac{1}{x} \, dx\)

\(= 3x^{\frac{1}{3}} \ln x - 3 \int x^{-\frac{2}{3}} \, dx\)

Integrating \(x^{-\frac{2}{3}}\), we get:

\(\int x^{-\frac{2}{3}} \, dx = \frac{x^{\frac{1}{3}}}{\frac{1}{3}} = 3x^{\frac{1}{3}}\)

Thus, the integral becomes:

\(3x^{\frac{1}{3}} \ln x - 9x^{\frac{1}{3}}\)

Evaluate from 1 to 8:

\(\left[ 3x^{\frac{1}{3}} \ln x - 9x^{\frac{1}{3}} \right]_{1}^{8}\)

\(= \left( 3 \cdot 8^{\frac{1}{3}} \ln 8 - 9 \cdot 8^{\frac{1}{3}} \right) - \left( 3 \cdot 1^{\frac{1}{3}} \ln 1 - 9 \cdot 1^{\frac{1}{3}} \right)\)

\(= \left( 6 \ln 8 - 18 \right) - \left( 0 - 9 \right)\)

\(= 6 \ln 8 - 18 + 9\)

\(= 6 \ln 8 - 9\)

Since \(\ln 8 = 3 \ln 2\), we have:

\(6 \ln 8 = 18 \ln 2\)

Thus, the final result is:

\(18 \ln 2 - 9\)