1. (a) Find \( \int \frac{\ln x}{x} \, dx \) - HSC - SSCE Mathematics Extension 2 - Question 1 - 2009 - Paper 1

For this integral, we apply integration by parts. Let:

• ( u = x ) and ( dv = e^{2x} , dx ), then ( du = dx ) and ( v = \frac{1}{2} e^{2x} ).

Now, we use the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substituting:

[ = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} , dx ] [ = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C ]

This integral can be approached by rewriting the expression:

[ = \int \frac{1}{4} \cdot \frac{4x^2}{1 + 4x^2} , dx + \int \frac{1}{1 + 4x^2} , dx ]

This gives us:

[ = \frac{1}{4} \cdot \int \frac{4u}{u} , du + \frac{1}{2} \tan^{-1}(2x) + C \text{ (where } u = 1 + 4x^2) ]

Therefore:

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

This integral is best approached using polynomial long division. First, divide ( x^5 - 6 ) by ( x^2 + 3x - 4 ). This will yield:

1. Perform the division: [ x^5 , / , (x^2 + 3x - 4) \text{ yields a polynomial } ]
2. Separate into simpler integrals: [ \int (\text{quotient}) , dx + \int \frac{\text{remainder}}{x^2 + 3x - 4} , dx ]
3. Solve each integral separately.

Finally, apply any necessary substitutions as part of the process.

First, simplify the integral:

[ \int \frac{\sqrt{3}}{2x^2 + 1} , dx ]

Now, recognize this as a standard integral form. We can use:

[ \int \frac{1}{ax^2 + b} , dx = \frac{1}{\sqrt{ab}} \tan^{-1}(\frac{\sqrt{a}}{b} x) + C ]

Substituting values:

[ = \frac{\sqrt{3}}{\sqrt{2}} \tan^{-1}(\frac{x}{\sqrt{2}}) + C ]