Partial Fractions

The method of partial fractions is based on the fact that a factorable rational function can be written as a sum of simpler fractions. Notice how evaluation of the following integral gives the answer as a sum of terms.

$\blacktriangleright$ Evaluate

$\dint$${\dfrac{{3x^{2}+2x+4}}{{(x-1)^{2}(x^{2}+1)^{2}}}}$dx = - ${\dfrac{{9}}{{4\left( x-1\right) }}}$ - ${\dfrac{{5}}{{2}}}$ln$\left(\vphantom{ x-1}\right.$x - 1$\left.\vphantom{ x-1}\right)$ + ${\dfrac{{5}}{{4}}}$ln$\left(\vphantom{ x^{2}+1}\right.$x2 + 1$\left.\vphantom{ x^{2}+1}\right)$ - ${\dfrac{{1}}{{4}}}$arctan x + ${\dfrac{{1}}{{8}}}$${\dfrac{{-4x-2}}{{% x^{2}+1}}}$

To gain an appreciation for how this calculation might be done internally, consider the method of partial fractions.

$\blacktriangleright$ To use the method of partial fractions on $\dint$${\dfrac{{3x^{2}+2x+4}}{{(x-1)^{2}(x^{2}+1)^{2}}}}$dx

1.
Enter the integral $\dint$${\dfrac{{3x^{2}+2x+4}}{{% (x-1)^{2}(x^{2}+1)^{2}}}}$dx.

2.
Select the rational expression ${\dfrac{{3x^{2}+2x+4}}{{% (x-1)^{2}(x^{2}+1)^{2}}}}$ inside the integral.

3.
While pressing the CTRL key, choose Partial Fractions from the Calculus submenu (or the Polynomials submenu).

4.
While the result is still selected, click itbpF0.3009in0.3009in0.0701inparens.wmfto get

$\displaystyle \dint$$\displaystyle \left(\vphantom{ \frac{9}{4\left( x-1\right) ^{2}}-\frac{5}{2\lef... ...frac{1+10x}{x^{2}+1}+\frac{1}{2}\frac{x-2}{\left( x^{2}+1\right) ^{2}}}\right.$$\displaystyle {\frac{{9}}{{4\left( x-1\right) ^{2}}}}$ - $\displaystyle {\frac{{5}}{{2\left( x-1\right) }}}$ + $\displaystyle {\frac{{1}}{{4}}}$$\displaystyle {\frac{{1+10x}}{{x^{2}+1}}}$ + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle {\frac{{x-2}}{{\left( x^{2}+1\right) ^{2}}}}$$\displaystyle \left.\vphantom{ \frac{9}{4\left( x-1\right) ^{2}}-\frac{5}{2\lef... ...frac{1+10x}{x^{2}+1}+\frac{1}{2}\frac{x-2}{\left( x^{2}+1\right) ^{2}}}\right)$dx

5.
The preceding integral can be written as a sum of four integrals.

$\displaystyle \dint$$\displaystyle {\frac{{9}}{{4\left( x-1\right) ^{2}}}}$dx - $\displaystyle \int$$\displaystyle {\frac{{5}}{{2\left( x-1\right) }% }}$dx + $\displaystyle {\frac{{1}}{{4}}}$$\displaystyle \int$$\displaystyle {\frac{{1+10x}}{{x^{2}+1}}}$dx + $\displaystyle {\frac{{1}}{{2}}}$$\displaystyle \int$$\displaystyle {\frac{{x-2}}{{\left( x^{2}+1\right) ^{2}}}}$dx

You can evaluate each of these integrals with Evaluate.

$\blacktriangleright$ Evaluate

$\dint$${\dfrac{{9}}{{4\left( x-1\right) ^{2}}}}$dx = - ${\dfrac{{9}}{{4\left( x-1\right) }}}$ 6pt

- $\dint$${\dfrac{{5}}{{2\left( x-1\right) }}}$dx = - ${\frac{{5}}{{2}}}$ln$\left(\vphantom{ 2x-2}\right.$2x - 2$\left.\vphantom{ 2x-2}\right)$ 6pt

${\dfrac{{1}}{{4}}}$$\dint$${\dfrac{{1+10x}}{{x^{2}+1}}}$dx = ${\dfrac{{5}}{{4}}}$ln$\left(\vphantom{ x^{2}+1}\right.$x2 + 1$\left.\vphantom{ x^{2}+1}\right)$ + ${\dfrac{{1}}{{4}}}$arctan x 6pt

${\dfrac{{1}}{{2}}}$$\dint$${\dfrac{{-2+x}}{{\left( x^{2}+1\right) ^{2}}}}$dx = ${\dfrac{{1}}{{8}}}$${\dfrac{{-4x-2}}{{x^{2}+1}}}$ - ${\dfrac{{1}}{{2}}}$arctan x

The original integral is the sum of the expressions on the right.


$\displaystyle \dint$$\displaystyle {\dfrac{{3x^{2}+2x+4}}{{(x-1)^{2}(x^{2}+1)^{2}}}}$dx = - $\displaystyle {\dfrac{{9}}{{4\left( x-1\right) }}}$ - $\displaystyle {\dfrac{{5}}{{2}}}$ln$\displaystyle \left(\vphantom{ 2x-2}\right.$2x - 2$\displaystyle \left.\vphantom{ 2x-2}\right)$  
    + $\displaystyle {\dfrac{{5}}{{4}}}$ln$\displaystyle \left(\vphantom{ x^{2}+1}\right.$x2 + 1$\displaystyle \left.\vphantom{ x^{2}+1}\right)$ - $\displaystyle {\dfrac{{1}}{{4}}}$arctan x + $\displaystyle {\dfrac{{1}}{{8}}}$$\displaystyle {\dfrac{{-4x-2}}{{x^{2}+1}}}$  

Applying Simplify to the difference between this answer and the earlier answer computed directly with Evaluate gives a constant.

- $\displaystyle {\frac{{5}}{{2}}}$ln$\displaystyle \left(\vphantom{ 2x-2}\right.$2x - 2$\displaystyle \left.\vphantom{ 2x-2}\right)$ + $\displaystyle {\frac{{5}}{{2}}}$ln$\displaystyle \left(\vphantom{ x-1}\right.$x - 1$\displaystyle \left.\vphantom{ x-1}\right)$ = - $\displaystyle {\frac{{5}}{{2}}}$ln 2