Find $\int \frac{\cos(\ln x)}{x^2} \, \mathrm{d}x$. Using the substitution $u = \sqrt{x+3}$, find $\int_1^6 \frac{x-2}{x\sqrt{x+3}} \, \mathrm{d}x$, giving your answer in the form
$a + \frac{b}{\sqrt{3}} \ln\left( \frac{c - \sqrt{d}}{c + \sqrt{d}} \right),$
where $a$, $b$, $c$ and $d$ are constants to be determined.

1. Apply integration by parts to the given integral.

Let $I = \int x^{-2} \cos(\ln x) \, \mathrm{d}x$.
Let $u = \cos(\ln x)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x^{-2}$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = -\frac{\sin(\ln x)}{x}$ and $v = -x^{-1}$.

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

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

Consider $\int x^{-2}\sin(\ln x) \, \mathrm{d}x$.
Let $u = \sin(\ln x)$ and $\frac{\mathrm{d}v}{\mathrm{d}x} = x^{-2}$.
Then $\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{\cos(\ln x)}{x}$ and $v = -x^{-1}$.

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

3. Substitute back and solve for $I$, adding the constant of integration.

Substitute the result from step 2 back into the equation for $I$:
$I = -\frac{\cos(\ln x)}{x} - \left(-\frac{\sin(\ln x)}{x} + \int x^{-2}\cos(\ln x) \, \mathrm{d}x \right)$
$I = -\frac{\cos(\ln x)}{x} + \frac{\sin(\ln x)}{x} - I$

Rearranging to solve for $I$:
$2I = \frac{\sin(\ln x) - \cos(\ln x)}{x}$
$I = \frac{\sin(\ln x) - \cos(\ln x)}{2x} + C$

1. Perform the substitution to find $\mathrm{d}x$ in terms of $\mathrm{d}u$ and find the new limits of integration.

$u = \sqrt{x+3} \implies u^2 = x + 3 \implies x = u^2 - 3$
Differentiating with respect to $u$:
$\frac{\mathrm{d}x}{\mathrm{d}u} = 2u \implies \mathrm{d}x = 2u \, \mathrm{d}u$

When $x = 1$, $u = \sqrt{1+3} = 2$.
When $x = 6$, $u = \sqrt{6+3} = 3$.

2. Substitute all $x$ terms and limits into the integral to express it entirely in terms of $u$.

$\int_1^6 \frac{x-2}{x\sqrt{x+3}} \, \mathrm{d}x = \int_2^3 \frac{(u^2 - 3) - 2}{(u^2 - 3)u} (2u) \, \mathrm{d}u$
$= \int_2^3 \frac{u^2 - 5}{u^2 - 3} \times 2 \, \mathrm{d}u$
$= \int_2^3 \frac{2u^2 - 10}{u^2 - 3} \, \mathrm{d}u$

3. Simplify the rational algebraic fraction by division to prepare for integration.

$\frac{2u^2 - 10}{u^2 - 3} = \frac{2(u^2 - 3) - 4}{u^2 - 3} = 2 - \frac{4}{u^2 - 3}$

$\text{So the integral becomes: } \int_2^3 \left( 2 - \frac{4}{u^2 - 3} \right) \, \mathrm{d}u$

4. Integrate using the standard integral formula for $\frac{1}{x^2 - a^2}$.

Using the standard result $\int \frac{1}{x^2-a^2} \, \mathrm{d}x = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right|$ with $a = \sqrt{3}$:

$= \left[ 2u - 4 \left( \frac{1}{2\sqrt{3}} \ln \left| \frac{u - \sqrt{3}}{u + \sqrt{3}} \right| \right) \right]_2^3$
$= \left[ 2u - \frac{2}{\sqrt{3}} \ln \left( \frac{u - \sqrt{3}}{u + \sqrt{3}} \right) \right]_2^3$

5. Substitute the limits to evaluate the definite integral.

$= \left( 6 - \frac{2}{\sqrt{3}} \ln \left( \frac{3 - \sqrt{3}}{3 + \sqrt{3}} \right) \right) - \left( 4 - \frac{2}{\sqrt{3}} \ln \left( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \right) \right)$
$= 2 + \frac{2}{\sqrt{3}} \left[ \ln \left( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \right) - \ln \left( \frac{3 - \sqrt{3}}{3 + \sqrt{3}} \right) \right]$

6. Combine the logarithmic terms and simplify the argument using rationalisation to reach the required form.

Combining the logarithms:
$= 2 + \frac{2}{\sqrt{3}} \ln \left( \frac{2 - \sqrt{3}}{2 + \sqrt{3}} \times \frac{3 + \sqrt{3}}{3 - \sqrt{3}} \right)$

Rationalising the fractions within the argument:
$\frac{2 - \sqrt{3}}{2 + \sqrt{3}} = \frac{(2 - \sqrt{3})^2}{4 - 3} = 7 - 4\sqrt{3}$
$\frac{3 + \sqrt{3}}{3 - \sqrt{3}} = \frac{(3 + \sqrt{3})^2}{9 - 3} = \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3}$

Multiplying them:
$(7 - 4\sqrt{3})(2 + \sqrt{3}) = 14 + 7\sqrt{3} - 8\sqrt{3} - 12 = 2 - \sqrt{3}$

We need to express $2 - \sqrt{3}$ in the form $\frac{c - \sqrt{d}}{c + \sqrt{d}}$. From the original upper limit evaluation, we can observe that:
$\frac{3 - \sqrt{3}}{3 + \sqrt{3}} = \frac{(3 - \sqrt{3})^2}{9 - 3} = \frac{12 - 6\sqrt{3}}{6} = 2 - \sqrt{3}$

Hence, substituting this back into the log term:
$= 2 + \frac{2}{\sqrt{3}} \ln \left( \frac{3 - \sqrt{3}}{3 + \sqrt{3}} \right)$

(This matches the target form with constants $a=2, b=2, c=3, d=3$.)