A calculus problem by Aman Joshi

$\begin{aligned} I&=\int_{0}^{1}\frac{3x^4+4x^3+3x^2}{\left(4x^3+3x^2+2x+1\right)^2}\,dx\\&=\int_{0}^{1}\frac{\frac{3}{x^2}+\frac{4}{x^3}+\frac{3}{x^4}}{\left(4+\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3}\right)^2}\,dx \\&=-\int_{\infty}^{10}\frac{1}{u^2}\,du & \small\color{#3D99F6}\text{where }u=4+\frac{3}{x}+\frac{2}{x^2}+\frac{1}{x^3} \\&=\frac{1}{10}\\k&=\boxed{10}\end{aligned}$

Thanks, @Aniket C-mmon for the solution.

Assuming that $\displaystyle \int \frac {P(x)}{(Q(x))^2} dx = \frac {P_1(x)}{Q(x)} + C$ , where $P_1(x)$ is a polynomial and $C$ is the constant of integration.

$\begin{aligned} \int \frac {3x^4+4x^3+3x^2}{(4x^3+3x^2+2x+1)^2} \ dx & = \frac {Ax^2+Bx+D}{4x^3+3x^2+2x+1} + C \\ \implies \frac {3x^4+4x^3+3x^2}{(4x^3+3x^2+2x+1)^2} & = \frac d{dx} \left(\frac {Ax^2+Bx+D}{4x^3+3x^2+2x+1}\right) \\ & = \frac {(2Ax+B)(4x^3+3x^2+2x+1) - (Ax^2+Bx+D)(12x^2+6x+2)}{(4x^3+3x^2+2x+1)^2} \\ & = \frac {-4Ax^4-8Bx^3+(2A-3B-12D)x^2 +(2A-6D)x+B-2D}{(4x^3+3x^2+2x+1)^2} \end{aligned}$

Comparing coefficients, we have $A = - \dfrac 34$ , $B= - \dfrac 12$ and $D= - \dfrac 14$ . Substituting in the equation,

$\begin{aligned} \int_0^1 \frac {3x^4+4x^3+3x^2}{(4x^3+3x^2+2x+1)^2} \ dx & = - \frac {3x^2+2x+1}{4(4x^3+3x^2+2x+1)} \ \bigg|_0^1 = - \frac 3{20} + \frac 14 = \frac 1{10} \end{aligned}$

$\implies k = \boxed{10}$

U can assume the given integration to be differentiation of a polynomial function divided by (1+2x+3x^2+4x^3) and then compare the co efficent to find the given polynomial