Question

Integration by Partial Fractions

1.) ∫(3x^2-x+1)/(x^3-x^2)dx

2.) ∫(t^2dt)/(t+1)^3

Answer 1

(1)

$∫\dfrac{(3x^2-x+1)}{(x^3-x^2)}dx\\[9pt]=\dfrac{3x^2-2x+x-1+2}{x^3-x^2}dx\\[9pt]=\dfrac{3x^2-2x}{x^3-x^2}dx+\dfrac{x-1}{x^2(x-1)}dx+\dfrac{2}{x^2(x-1)}dx\\[9pt]$

Let $x^3-x^2=t$

$\Rightarrow (3x^2-2x)dx=dt$ and using partial fraction in the third term-

$=\dfrac{dt}{t}+\dfrac{1}{x^2}dx+\dfrac{2}{x-1}dx-\dfrac{2}{x}dx-\dfrac{2}{x^2}dx\\[9pt]=lnt-\dfrac{1}{x}+2ln(x-1)-2lnx+\dfrac{2}{x}+C\\[9pt]=ln(x^3-x^2)+\dfrac{1}{x}+2ln\dfrac{x-1}{x}+C$

(2)

$∫\dfrac{(t^2)}{(t+1)^3}dt\\[9pt]=\dfrac{(t+1-1)^2}{(t+1)^3}dt\\[9pt]=\dfrac{(t+1)^2+1-2(t+1)}{(t+1)^3}dt\\[9pt]=\dfrac{1}{t+1}dt+\dfrac{1}{(t+1)^3}dt-\dfrac{2}{(t+1)^2}dt\\[9pt]=ln(t+1)-\dfrac{1}{2(t+1)^2}+\dfrac{2}{(t+1)}+C$