Find $$\int_{0}^{3} \tan^3{x} \sec^2{x} \, dx.$$ (b) By completing the square, find $$\int \frac{dx}{\sqrt{-4+x^{2}+1}}.$$ (c) Use integration by parts to evaluate $$\int \frac{\ln{x}}{x^{2}} \, dx.$$ (d) Use the substitution $u = \sqrt{x-1}$ to evaluate $$\int_{2}^{3} \frac{1+x}{\sqrt{1}} \, dx.$$ (e) (i) Find real numbers $\alpha$ and $b$ such that $$\frac{5x^{2}-3x+1}{(x+1)(x-2)} = \frac{\alpha x + 1}{x+1} + \frac{b}{x-2}.$$ (ii) Find $$\int \frac{5x^{2}-3x+1}{(x^{2}+1)(x-2)} \, dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2001 - Paper 1

To complete the square for the expression (-4 + x^{2} + 1), we get:

$$\sqrt{x^2 - 3} = \sqrt{(x - \sqrt{3})(x + \sqrt{3})}.$$

The integral becomes: $\int \frac{dx}{\sqrt{(x - \sqrt{3})(x + \sqrt{3})}}$ To evaluate this integral, we can use trigonometric or hyperbolic substitution based on the completed square.

For this integral, we will use the integration by parts formula: $\int u \, dv = uv - \int v \, du$. Let:

• ( u = \ln{x} \Rightarrow du = \frac{1}{x} , dx )
• ( dv = \frac{1}{x^2} , dx \Rightarrow v = -\frac{1}{x} ) Then we can substitute into the parts formula:

$\int \frac{\ln{x}}{x^{2}} \, dx = -\frac{\ln{x}}{x} - \int -\frac{1}{x} \left(-\frac{1}{x}\right) \, dx$ This simplifies to: $\int \frac{\ln{x}}{x^{2}} \, dx = -\frac{\ln{x}}{x} + \int \frac{1}{x^2} \, dx$

Starting with the substitution: Let ( u = \sqrt{x-1} \Rightarrow x = u^2 + 1 \Rightarrow dx = 2u , du ). We also need to change the bounds:

• When ( x = 2, u = 1 )
• When ( x = 3, u = \sqrt{2} ) Substituting these into the integral, we have:

$\int_{1}^{\sqrt{2}} \frac{1 + (u^2 + 1)}{\sqrt{u^2}} \cdot 2u \, du$ Which simplifies to an integral that can be evaluated.

To find the values of ( \alpha ) and ( b ), we can assume:

$5x^{2}-3x+1 = (\alpha x + 1)(x-2) + b(x+1).$ By expanding both sides and equating coefficients, we will find a system of equations to solve for ( \alpha ) and ( b ).

To evaluate this integral, we will first perform partial fraction decomposition on the expression:

$\frac{5x^{2}-3x+1}{(x^{2}+1)(x-2)}$. Next, we will express this as: $A \cdot \frac{1}{x^{2}+1} + B \cdot \frac{1}{x-2}$ We will determine the constants ( A ) and ( B ), then integrate each term separately.