Question

    X^3+3X^2+X+9
      /  /   /      dx
    (X^2+1)(X^2+3)
    
    
     

Answer 1

$\begin{aligned} &\int\dfrac{x^3+3x^2+x+9}{ (x^2+1)(x^2+3)} dx = \\ \\ &\textsf{performing partial fraction decomposition}\\ &\dfrac{x^3+3x^2+x+9}{ (x^2+1)(x^2+3)}= \dfrac{A}{(x²+1)} + \dfrac{B}{(x²+3)} \\ &\dfrac{x}{(x²+3)} + \dfrac{3}{(x²+1)} \\ \\ \\ &\therefore \int\dfrac{x^3+3x^2+x+9}{ (x^2+1)(x^2+3)} dx = \int\dfrac{x}{(x²+3)} dx+ \int \dfrac{3}{(x²+1)}dx \end{aligned}$

for,

$\int \dfrac{x}{x²+3} dx \\$

let u = x² + 3, du = 2xdx

$\dfrac12\int\dfrac1udu = \dfrac{\ln u}2 + c = \dfrac12\ln (x²+3) + c$

for,

$\int \dfrac{3}{x²+1} dx = 3\int \dfrac{1}{x²+1} dx$

but, $\int \dfrac{1}{x²+1} dx = \tan^{-1}x \\$

$\int \dfrac{3}{x²+1} dx = 3\tan^{-1}x + c$

$\begin{aligned}\therefore\int\dfrac{x^3+3x^2+x+9}{ (x^2+1)(x^2+3)} dx = \dfrac12\ln (x²+3) + 3\tan^{-1}x + c \end{aligned}$