Answer to Question #186226 in Calculus for Phyroe

Question

Integration by Partial Fractions

1.) ∫dz/z+z^3

2.) ∫(2s+1)ds/s^2(s^2+1)

Answer 1

1.) $\int \dfrac{dz}{z+z^3}$

Partial fraction can be written as,

$f(z) = \dfrac{1}{z} - \dfrac{z}{z^2+1}$

$f(z) = \dfrac{1}{z} - \dfrac{1}{2} \dfrac{1}{(z+i)}+ \dfrac{1}{2} \dfrac{1}{(z-i)}$

Integrating both sides,

$\int f(z)dz = logz - \dfrac{1}{2}log(z+i) + \dfrac{1}{2}log(z-i) + logC$

b.) $\int \dfrac{2x+1}{x^2(x^2+1)} dx$

Performing partial fraction decomposition:

$\dfrac{2x+1}{x^2(x^2+1)} = \dfrac{A}{x} + \dfrac{B}{x^2}+ \dfrac{Cx+D}{x^2+1}$

$2x+1 = x^2(Cx+D) + x(x^2+1)A + (x^2+1)B$

$2x+1 = x^3(A+C) + x^2(B+D) + xA+B$

After equating coefficients we get,

$A= 2 , B=1 ,C = -2 \hspace{2mm}and \hspace{2mm} D = -1$

Hence,

$= \int (\dfrac{-2x-1}{x^2+1}+ \dfrac{2}{x}+ \dfrac{1}{x^2})dx$

Apply linearity:

$= -\int \dfrac{2x+1}{x^2+1}dx + 2\int \dfrac{1}{x}dx + \int \dfrac{1}{x^2}dx$

$= -\int \dfrac{2x}{x^2+1}dx - \int \dfrac{1}{x^2+1}dx + 2\int \dfrac{1}{x}dx + \int \dfrac{1}{x^2}dx$

$= -log(x^2+1) - tan^{-1}x + 2logx - \dfrac{1}{x}+ C$