Find $$rac{ ext{ln} x}{x} \, dx.$$ Find $$ ext{e}^{2x} \, dx.$$ Find $$rac{x^2}{1 + 4x^2} \, dx.$$ Evaluate $$rac{ ext{5}}{2} rac{x - 6}{x^2 + 3x - 4} \, dx.$$ Evaluate $$rac{ ext{3}}{1 + x^2} \, dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2009 - Paper 1

For the integral ( \int \text{e}^{2x} , dx ), we can use substitution. Let ( u = 2x ), then ( du = 2 , dx ) or ( dx = \frac{du}{2} ). This changes the integral to ( \frac{1}{2} \int \text{e}^u , du = \frac{1}{2} \text{e}^u + C = \frac{1}{2} \text{e}^{2x} + C ).

To evaluate the integral, we rewrite it as ( \int \frac{x^2}{1 + 4x^2} , dx = \int \left( \frac{1}{4} - \frac{1}{4(1 + 4x^2)} \right) , dx ). The first term equals ( \frac{1}{4}x ) and the second term can be integrated as follows: ( -\frac{1}{4} \int \frac{1}{1 + 4x^2} , dx = -\frac{1}{8} \text{arctan}(2x) + C ). Thus, the final result is: ( \frac{x}{4} - \frac{1}{8} \text{arctan}(2x) + C ).

We begin by factoring the denominator: ( x^2 + 3x - 4 = (x + 4)(x - 1) ). This allows us to perform partial fraction decomposition: ( \frac{5}{(x + 4)(x - 1)} = \frac{A}{x + 4} + \frac{B}{x - 1} ). Solving gives: ( A = 1, B = 4 ). Now integrating: ( \int \left( \frac{1}{x + 4} + \frac{4}{x - 1} \right) dx = \ln |x + 4| + 4 \ln |x - 1| + C ).

This integral is straightforward since ( \int \frac{1}{1 + x^2} , dx = \text{arctan}(x) + C ). Hence, multiplying by 3 gives: ( 3 \int \frac{1}{1 + x^2} , dx = 3 \text{arctan}(x) + C ).