Answer to Question #185639 in Calculus for phyroe

Question

Integration by Parts Fractions

1.) ∫(3x^2-x+1)/(x^3-x^2)dx

2.) ∫(t^2dt)/(t+1)^3

Answer 1

Solution:

1.

${\displaystyle\int}\dfrac{3x^2-x+1}{x^3-x^2}\,\mathrm{d}x$

Factor the denominator:

${\displaystyle\int}\dfrac{3x^2-x+1}{\left(x-1\right)x^2}\,\mathrm{d}x$

$={\displaystyle\int}\left(\dfrac{3}{x-1}-\dfrac{1}{x^2}\right)\mathrm{d}x\\$

$={3}{\displaystyle\int}\dfrac{1}{x-1}\,\mathrm{d}x-{\displaystyle\int}\dfrac{1}{x^2}\,\mathrm{d}x$

Now solving these two integration separately

 

Substitute u=x-1 

${\displaystyle\int}\dfrac{1}{x-1}\,\mathrm{d}x={\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u=\ln\left(u\right)=\ln\left(x-1\right)$

${\displaystyle\int}\dfrac{1}{x^2}\,\mathrm{d}x=-\dfrac{1}{x}$

$\therefore \ {3}{\displaystyle\int}\dfrac{1}{x-1}\,\mathrm{d}x-{\displaystyle\int}\dfrac{1}{x^2}\,\mathrm{d}x=3\ln\left(x-1\right)+\dfrac{1}{x}$

Apply the absolute value function to arguments of logarithm functions in order to extend the anti derivative's domain:

${\displaystyle\int}\dfrac{3x^2-x+1}{x^3-x^2}\,\mathrm{d}x=\dfrac{1}{x}+3\ln\left(\left|x-1\right|\right)+C$

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2.

${\displaystyle\int}\dfrac{t^2}{\left(t+1\right)^3}\,\mathrm{d}t \\ Substitute \ u=t+1 \longrightarrow \dfrac{\mathrm{d}u}{\mathrm{d}t} = 1 \longrightarrow \ dt=du$

$={\displaystyle\int}\dfrac{\left(u-1\right)^2}{u^3}\,\mathrm{d}u \\\newline ={\displaystyle\int}\left(\dfrac{1}{u}-\dfrac{2}{u^2}+\dfrac{1}{u^3}\right)\mathrm{d}u \\ ={\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u-{2}{\displaystyle\int}\dfrac{1}{u^2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{u^3}\,\mathrm{d}u \\$

Now solving each integration separately

${\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u=\ln\left(u\right)\\{\displaystyle\int}\dfrac{1}{u^2}\,\mathrm{d}u=-\dfrac{1}{u}\\{\displaystyle\int}\dfrac{1}{u^3}\,\mathrm{d}u=-\dfrac{1}{2u^2}$

$\therefore \ {\displaystyle\int}\dfrac{1}{u}\,\mathrm{d}u-{2}{\displaystyle\int}\dfrac{1}{u^2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{u^3}\,\mathrm{d}u=\ln\left(u\right)+\dfrac{2}{u}-\dfrac{1}{2u^2}$

re substituting u=t+1

$=\ln\left(t+1\right)+\dfrac{2}{t+1}-\dfrac{1}{2\left(t+1\right)^2}$

Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivative's domain:

${\displaystyle\int}\dfrac{t^2}{\left(t+1\right)^3}\,\mathrm{d}t=\ln\left(\left|t+1\right|\right)+\dfrac{2}{t+1}-\dfrac{1}{2\left(t+1\right)^2}+C$

It can be further simplified to 

${\displaystyle\int}\dfrac{t^2}{\left(t+1\right)^3}\,\mathrm{d}t \ = \ \ln\left(\left|t+1\right|\right)+\dfrac{4t+3}{2t^2+4t+2}+C$