Find

1. Rewrite the integral to the form $k \int \frac{f'(x)}{f(x)} \,\mathrm{d}x$.

Let $f(x) = 5 - 2\mathrm{e}^x$. Then $f'(x) = -2\mathrm{e}^x$.

$\int \frac{10\mathrm{e}^x}{5 - 2\mathrm{e}^x} \,\mathrm{d}x = -5 \int \frac{-2\mathrm{e}^x}{5 - 2\mathrm{e}^x} \,\mathrm{d}x$

2. Integrate to obtain the natural logarithm.

$= -5 \ln |5 - 2\mathrm{e}^x| + c$

1. Rewrite the integral in the form $k \int \mathrm{f}'(x)[\mathrm{f}(x)]^n \,\mathrm{d}x$ by adjusting the constant.

$\int \frac{x}{\sqrt{1 + 8x^2}} \,\mathrm{d}x = \int x(1 + 8x^2)^{-\frac{1}{2}} \,\mathrm{d}x$
$= \frac{1}{16} \int 16x(1 + 8x^2)^{-\frac{1}{2}} \,\mathrm{d}x$

2. Integrate using the reverse chain rule and simplify.

$= \frac{1}{16} \left[ \frac{(1 + 8x^2)^{\frac{1}{2}}}{\frac{1}{2}} \right] + c$
$= \frac{1}{8}\sqrt{1 + 8x^2} + c$

1. Set up the first integration by parts

Let $u = (\ln x)^2$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = 2(\ln x) \cdot \frac{1}{x} = \frac{2\ln x}{x}$ and $v = \frac{x^2}{2}$.

Using $\int u \frac{\mathrm{d}v}{\mathrm{d}x} \,\mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \,\mathrm{d}x$,

$\int x(\ln x)^2 \,\mathrm{d}x = \frac{x^2}{2}(\ln x)^2 - \int \frac{x^2}{2} \left(\frac{2\ln x}{x}\right) \,\mathrm{d}x$

2. Simplify the remaining integral

$\int x(\ln x)^2 \,\mathrm{d}x = \frac{x^2}{2}(\ln x)^2 - \int x \ln x \,\mathrm{d}x$

3. Apply integration by parts a second time to the new integral

For $\int x \ln x \,\mathrm{d}x$, let $u_1 = \ln x$ and $\frac{\mathrm{d}v_1}{\mathrm{d}x} = x$.
Then $\frac{\mathrm{d}u_1}{\mathrm{d}x} = \frac{1}{x}$ and $v_1 = \frac{x^2}{2}$.

$\int x \ln x \,\mathrm{d}x = \frac{x^2}{2}\ln x - \int \frac{x^2}{2} \left(\frac{1}{x}\right) \,\mathrm{d}x$
$= \frac{x^2}{2}\ln x - \int \frac{x}{2} \,\mathrm{d}x$

4. Integrate the final term, combine the results, and add the constant of integration

$\int x \ln x \,\mathrm{d}x = \frac{x^2}{2}\ln x - \frac{x^2}{4}$

Substituting this back into the first equation:
$\int x(\ln x)^2 \,\mathrm{d}x = \frac{x^2}{2}(\ln x)^2 - \left(\frac{x^2}{2}\ln x - \frac{x^2}{4}\right) + c$
$= \frac{x^2}{2}(\ln x)^2 - \frac{x^2}{2}\ln x + \frac{x^2}{4} + c$