Evaluate $$egin{align*} ext{(a)} \ ext{Evaluate } \ ext{ } \int_{0}^{1} \frac{e^x}{(1+e^x)^2}dx - HSC - SSCE Mathematics Extension 2 - Question 1 - 2003 - Paper 1

Let:

• $u = \log_e x \Rightarrow du = \frac{1}{x} dx$
• $dv = x^3 dx \Rightarrow v = \frac{x^4}{4}$

Applying integration by parts:

$\int x^3 \log_e x \, dx = uv - \int v \, du = \frac{x^4}{4} \log_e x - \int \frac{x^4}{4} \cdot \frac{1}{x} dx = \frac{x^4}{4} \log_e x - \frac{1}{4} \int x^3 dx$

Continuing:

$= \frac{x^4}{4} \log_e x - \frac{1}{4} \cdot \frac{x^4}{4} = \frac{x^4}{4} \log_e x - \frac{x^4}{16} + C$

To evaluate this integral, we first complete the square:

$\int \frac{dx}{\sqrt{2x-5}} = \int \frac{1}{\sqrt{2(x - \frac{5}{2})}} \, dx$

Using substitution $u = \sqrt{2} \sqrt{x - \frac{5}{2}}$, we can transform the integral to:

$= \frac{1}{\sqrt{2}} \int \frac{du}{u} = \frac{1}{\sqrt{2}} \log_e |u| + C$

Substituting back in for $u$ gives:

$= \frac{1}{\sqrt{2}} \log_e |\sqrt{2} \sqrt{x - \frac{5}{2}}| + C$

Start by multiplying both sides by $(x-1)(x^2 + 4)$:

$5x^2 - 3x + 13 = a(x^2 + 4) + (bx - 1)(x - 1)$

Expanding the right-hand side, we get:

$5x^2 - 3x + 13 = ax^2 + 4a + bx^2 - bx - x + 1$

Combining like terms gives:

$(a + b)x^2 + (4a - b - 1)x + (4a + 1) = 5x^2 - 3x + 13$

From which we can derive the equations:

• $a + b = 5$
• $4a - b - 1 = -3$
• $4a + 1 = 13$

Solving these, we find:

• $a = 3, b = 2$.

Using the results from part (d), we have:

$\int \frac{3}{x-1} \, dx + \int \frac{2x-1}{x^2+4} \, dx$

The first integral evaluates to:

$3 \log_e |x-1| + C_1$

The second integral requires completing the square or recognizing it as a standard integral:

$\int \frac{2x}{x^2 + 4} \, dx - \int \frac{1}{x^2 + 4} \, dx$

Thus,

$\frac{1}{2} \log_e |x^2 + 4| - \frac{1}{2} \tan^{-1}\left( \frac{x}{2} \right) + C_2$

Combining, the overall answer is:

$3 \log_e |x-1| + \frac{1}{2} \log_e |x^2 + 4| - \frac{1}{2} \tan^{-1}\left( \frac{x}{2} \right) + C$

Using the substitution $x = 3\sin \theta$, we find:

$dx = 3\cos \theta \, d\theta$

The limits transform to:

• When $x = 0$, $\theta = 0$.
• When $x = \frac{3}{2}$, $\theta = \sin^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{6}$.

The integral becomes:

$\int_{0}^{\frac{\pi}{6}} \frac{3\cos\theta \, d\theta}{(9 - 9\sin^2\theta)^{\frac{3}{2}}} = \int_{0}^{\frac{\pi}{6}} \frac{3\cos\theta \, d\theta}{(9(\cos^2\theta))^{\frac{3}{2}}} = \int_{0}^{\frac{\pi}{6}} \frac{3\cos\theta \, d\theta}{27\cos^3\theta} = \frac{1}{9} \int_{0}^{\frac{\pi}{6}} \sec^2\theta \, d\theta$

This evaluates to:

$\frac{1}{9} \tan \theta \Bigg|_{0}^{\frac{\pi}{6}} = \frac{1}{9} \left( \tan\left( \frac{\pi}{6} \right) - \tan(0) \right) = \frac{1}{9} \left( \frac{1}{\sqrt{3}} - 0 \right) = \frac{1}{9\sqrt{3}}$