Let $f(x) = 3x^2 + x$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2001 - Paper 1

To solve this integral, we can use substitution. Let:

$u = 1 + e^x,$ thus $du = e^x dx.$

Now, we rewrite the integral in terms of $u$:

$\int \frac{e^x}{1 + e^x} dx = \int \frac{du}{u} = \ln|u| + C = \ln(1 + e^x) + C.$

To calculate this integral, we first find the antiderivative:

$\int 3 \cos(2x)dx = \frac{3}{2} \sin(2x) + C.$

Now, we evaluate from $0$ to $c$:

$\int_0^c 3 \cos(2x)dx = \left[ \frac{3}{2} \sin(2x) \right]_0^c = \frac{3}{2} \sin(2c) - \frac{3}{2} \sin(0) = \frac{3}{2} \sin(2c).$

The word ALPHABRIC has a total of 9 letters, with the letter A appearing twice. The number of arrangements is calculated using:

$\frac{9!}{2!} = \frac{362880}{2} = 181440.$

The vowels in ALPHABRIC are A, A, I. The number of ways to arrange them in the specified positions is:

$\frac{3!}{2!} = 3.$

For the remaining letters (L, P, H, B, R, C) which must fill the other 5 positions:

$6! = 720.$

Therefore, the total arrangements are:

$3 \times 720 = 2160.$

To find the term independent of $x$, we set the exponent of $x$ to zero in the expansion:

Using the binomial theorem:

$T_k = \binom{9}{k} (x^2)^{9-k} \left(-\frac{1}{x}\right)^k = \binom{9}{k} (-1)^k x^{2(9-k) - k} = \binom{9}{k} (-1)^k x^{18 - 3k}.$

To find the term independent of $x$, set $18 - 3k = 0$:

$k = 6.$

The term is:

$T_6 = \binom{9}{6} (-1)^6 = 84.$