Question

3.1.4 ∫(6x - 10)/((x² - 6x + 9)(x + 1)) dx use partial fractions (9)
3.2 a force f stretches a spring through a distance x. the spring obeys hooke’s law, namely f = -kx, where k is the spring constant. the work done, w, is given by w = ∫f dx. determine the work, w, if a spring, with spring constant 30 n/m is stretched by 40 cm. (5)

Explanation:

Step1: Decompose into partial - fractions for 3.1.4

We have $\frac{6x - 10}{(x^{2}-6x + 9)(x + 1)}=\frac{6x - 10}{(x - 3)^{2}(x + 1)}$. Let $\frac{6x - 10}{(x - 3)^{2}(x + 1)}=\frac{A}{x+1}+\frac{B}{x - 3}+\frac{C}{(x - 3)^{2}}$. Then $6x-10=A(x - 3)^{2}+B(x + 1)(x - 3)+C(x + 1)$.
If $x=-1$, then $6(-1)-10=A(-1 - 3)^{2}$, so $-16 = 16A$, and $A=-1$.
If $x = 3$, then $6\times3-10=C(3 + 1)$, so $8 = 4C$, and $C = 2$.
Comparing the coefficients of $x^{2}$: $0=A + B$, since $A=-1$, then $B = 1$.
So $\frac{6x - 10}{(x - 3)^{2}(x + 1)}=-\frac{1}{x+1}+\frac{1}{x - 3}+\frac{2}{(x - 3)^{2}}$.
Then $\int\frac{6x - 10}{(x^{2}-6x + 9)(x + 1)}dx=-\int\frac{1}{x+1}dx+\int\frac{1}{x - 3}dx+2\int\frac{1}{(x - 3)^{2}}dx$.
$=-\ln|x + 1|+\ln|x - 3|-\frac{2}{x - 3}+C=\ln|\frac{x - 3}{x + 1}|-\frac{2}{x - 3}+C$.

Step2: Calculate work for 3.2

First, convert the distance to SI units. $x = 40\ cm=0.4\ m$ and $k = 30\ N/m$.
Since $F=-kx$, then $W=\int_{0}^{0.4}-kx\ dx$.
Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)$, we have $W=-k\int_{0}^{0.4}x\ dx=-30\times[\frac{x^{2}}{2}]_{0}^{0.4}$.
$=-30\times\frac{0.4^{2}}{2}=-30\times0.08=-2.4\ J$. The negative sign just indicates the direction of the force relative to the displacement. The magnitude of the work done is $2.4\ J$.

Answer:

3.1.4: $\ln|\frac{x - 3}{x + 1}|-\frac{2}{x - 3}+C$
3.2: $2.4\ J$