Find $$ \int x \ln x \, dx $$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2011 - Paper 1

This integral can be solved by substitution. Let:

• $x = \tan \theta$, then $dx = \sec^2 \theta \, d\theta$.
• The limits change to $0$ to rac{\pi}{4} as $x$ goes from $0$ to $3$.

So, we have:

$\int \sqrt{x^2 + 1} \, dx = \int \sqrt{\tan^2 \theta + 1} \cdot \sec^2 \theta \, d\theta = \int \sec^3 \theta \, d\theta$.

This is a standard integral:

$\int \sec^3 \theta \, d\theta = \frac{1}{2}(\sec \theta \tan \theta + \ln | \sec \theta + \tan \theta |) + C$.

To find $a$, $b$, and $c$, multiply through by the common denominator $x^2(x-1)$:

$1 = a x (x-1) + b (x-1) + c x^2$.

Expanding and rearranging:

$1 = (a + c)x^2 + (-a + b)x - b$.

Setting up the equations:

1. $a + c = 0$
2. $-a + b = 0$
3. $-b = 1$

Solving gives $b = -1$, $a = -1$, and $c = 1$.

Using the values found:

$\frac{1}{x^2(x-1)} = \frac{-1}{x} + \frac{-1}{x^2} + \frac{1}{x-1}$.

Integrating each term separately:

$-\int \frac{1}{x} \, dx - \int \frac{1}{x^2} \, dx + \int \frac{1}{x-1} \, dx = -\ln |x| + \frac{1}{x} + \ln |x-1| + C$.

To solve this integral, we can use the identity:

$\cos^3 \theta = \cos \theta (1 - \sin^2 \theta)$.

This gives:

$\int \cos^3 \theta \, d\theta = \int \cos \theta \, d\theta - \int \cos \theta \sin^2 \theta \, d\theta$.

The first integral is:

$\int \cos \theta \, d\theta = \sin \theta$,

The second can be solved using substitution: Let $u = \sin \theta$, then $du = \cos \theta \, d\theta$:

$-\int u^2 \, du = -\frac{u^3}{3} = -\frac{\sin^3 \theta}{3}$.

Thus, combining gives:

$\int \cos^3 \theta \, d\theta = \sin \theta - \frac{\sin^3 \theta}{3} + C$.

This integral can be solved with a simple substitution. The denominator can be factored or evaluated with the observation of symmetry:

Using the substitution $u = 5 - 2t^2$, the limits change accordingly:

Now, we can express it in simpler form and recognize that it's an even function:

Thus,

the solution can be derived as:

= rac{1}{\sqrt{10}} [\tan^{-1} (\frac{t}{\sqrt{5/2}})]_{-1}^{1},

which evaluates the bounds to give a final answer.