Q.(a) Evaluate : $\int \dfrac{x}{2x^4 - 3x^2 - 2}\, dx$.

(a) With $u=x^2$ the integral reduces to partial fractions, giving $\tfrac{1}{10}\log\left|\tfrac{x^2-2}{2x^2+1}\right| + c$. (b) The homogeneous DE integrates (via $y = vx$) to $x^2 y(2x+y) = k$, and $k = 768$.

Part (a) — Integration

Let $u = x^2 \Rightarrow du = 2x\,dx$, so $x\,dx = \tfrac12\,du$:

$\int \frac{x\,dx}{2x^4 - 3x^2 - 2} = \frac12\int \frac{du}{2u^2 - 3u - 2}.$

Factor: $2u^2 - 3u - 2 = (2u + 1)(u - 2)$. Partial fractions:

$\frac{1}{(2u+1)(u-2)} = \frac{A}{2u+1} + \frac{B}{u-2}.$

$1 = A(u-2) + B(2u+1)$. At $u = 2$: $B = \tfrac15$; at $u = -\tfrac12$: $A = -\tfrac25$.

$\frac12\int\left[\frac{-2/5}{2u+1} + \frac{1/5}{u-2}\right]du = \frac12\left[-\frac{2}{5}\cdot\frac12\log|2u+1| + \frac{1}{5}\log|u-2|\right] + c.$

$= \frac{1}{10}\big(\log|u-2| - \log|2u+1|\big) + c = \frac{1}{10}\log\left|\frac{x^2 - 2}{2x^2 + 1}\right| + c.$

Part (b) — Total cost function

DE: $(3xy + y^2)\,dx + (x^2 + xy)\,dy = 0$ — homogeneous (each term degree $2$). Put $y = vx$, $dy = v\,dx + x\,dv$; divide by $x^2$:

$(3v + v^2)\,dx + (1 + v)(v\,dx + x\,dv) = 0.$

$(4v + 2v^2)\,dx + (1 + v)x\,dv = 0 \ \Rightarrow\ \frac{dx}{x} = -\frac{(1+v)}{2v(v+2)}\,dv.$

Partial fractions: $\dfrac{1+v}{2v(v+2)} = \dfrac{1/4}{v} + \dfrac{1/4}{v+2}$. Integrating: …