Question

integrate x2 +x2 +5/(x2+4) (x+1) dx

Answer 1

I = $\int$ (x² + x² + $\frac{5}{(x^2+4)(x+1)}$ )dx

Converting in partial fractions form, we get-

$\frac{5}{(x^2+4)(x+1)} = \frac{A}{(x+1)}+\frac{Bx+C}{(x^2+4)}$

5 = (A+B)$x^2$ + (B+C)x + (4A+C)

Equating the coefficients and solving equations, we get

A = 1 ; B = -1 ; C = 1

I = $\int$ (x² + x² + $\frac{1}{(x+1)}-\frac{x-1}{x^2+4}$ )dx

I = $\int$ (2x² + $\frac{1}{(x+1)}-\frac{x}{(x^2+4)}+\frac{1}{(x^2+4)}$ )dx

Since, integration of $\frac{1}{x+1}$ = ln(x+1) , integration of $\frac{x}{x^2+4} is \frac{1}{2}ln(x^2+4)$ and integration of $\frac{1}{x^2+4}$ is $\frac{1}{2}tan^{-1}\frac{x}{2}$ , we have

$I = \frac{2x^3}{3}+ln(x+1)-\frac{1}{2}[ln(x^2+4)-tan^{-1}(\frac{x}{2})]+const$