Question

Integrate ∫x^2+x+5/(x^2+4)(x+1)dx

Answer 1

$\intop \frac{x^2+x+5}{(x^2+4)(x+1)} dx$

Let's decompose a fraction:

$\frac{x^2+x+5}{(x^2+4)(x+1)} = \frac{Kx+L}{x^2+4} + \frac{M}{x+1} = \frac{(Kx+L)(x+1)+M(x^2+4)}{(x^2+4)(x+1)} = \frac{Kx^2+Kx+Lx+L+Mx^2+4M}{(x^2+4)(x+1)} = \frac{(K+M)x^2+(K+L)x+(L+4M)}{(x^2+4)(x+1)}$

$\begin{cases} K+M=1 \\ K+L=1 \\ L+4M=5 \end{cases}$

$\begin{cases} M=1-K \\ L=1-K \\ L+4M=5 \end{cases}$

$1-K+4(1-K)=5$

$1-K+4-4K=5$

$-5K=0$

$K=0$

$M=1-K=1-0=1$

$L=1-K=1-0=1$

So,

$\intop \frac{x^2+x+5}{(x^2+4)(x+1)} dx = \intop (\frac{0x+1}{x^2+4} + \frac{1}{x+1})dx = \intop (\frac{1}{x^2+4} + \frac{1}{x+1})dx = \intop \frac{dx}{x^2+4} + \intop \frac{dx}{x+1} = I_1 + I_2$

$I_1 = \intop \frac{dx}{x^2+4} = \intop \frac{dx}{x^2+2^2} = \frac{1}{2} \arctan \frac{x}{2} +C$

$I_2 = \intop \frac{dx}{x+1} = \ln (x+1) + C$

$\intop \frac{x^2+x+5}{(x^2+4)(x+1)} dx = \frac{1}{2} \arctan \frac{x}{2} + \ln (x+1) + C$