Answer to Question #186223 in Calculus for Phyroe

Question

Integration by Partial Fractions

1.) ∫dy/(y^2+2y)

2.) ∫(x^3+x+2)/(x^2-1)dx

Answer 1

(a)

$\int \frac{dy}{y^2+2y}\newline =\int \frac{dy}{y(2+y)}$

By partial fraction,

$\frac{1}{y(2+y)}=\frac{A}{y}+\frac{B}{2+y} \newline \text{by solving, we get A=1/2 and B= -1/2}$

$\int \frac{dy}{y^2+2y}\newline =\frac{1}{2}\int (\frac{1}{y}-\frac{1}{2+y})dy\newline =\frac{1}{2}[ logy-log(2+y)]\newline =\frac{1}{2} log\frac{y}{(2+y)}$

(b)

By using partial fraction and algebraic indentities,

$\frac{x^3+x+2}{x^2-1}=\frac{(x+1)(x^2-x+2)}{(x+1)(x-1)}=\frac{x^2-x+2}{x-1}=x+\frac{2}{x-1} \newline$

Then, the integral

$\int \frac{x^3+x+2}{x^2-1}dx=\int (x+\frac{2}{x-1} )dx\newline =x^2+2log(x-1)$