# Find the form of the partial fraction decomposition of: 1. x^4+6 P

Find the form of the partial fraction decomposition of:
1. $$\displaystyle{x}^{{4}}+{6}$$
$$\displaystyle{x}^{{5}}+{7}{x}^{{3}}$$
2. 2 $$\displaystyle{\left({x}^{{2}}-{9}\right)}^{{2}}$$

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SoosteethicU
1. $$\displaystyle\frac{{{x}^{{4}}+{6}}}{{{x}^{{5}}+{7}{x}^{{3}}}}=\frac{{{x}^{{4}}+{6}}}{{{x}^{{3}}{\left({x}^{{2}}+{7}\right)}}}$$
The partial fraction:
$$\displaystyle=\frac{{A}}{{x}}+\frac{{B}}{{x}^{{2}}}+\frac{{C}}{{x}^{{3}}}+\frac{{{D}{x}+{E}}}{{{x}^{{2}}+{7}}}$$
Where A-E are constants.
2. $$\displaystyle\frac{{2}}{{{\left({x}^{{2}}-{9}\right)}^{{2}}}}=\frac{{2}}{{{\left({\left({x}+{3}\right)}{\left({x}-{3}\right)}\right)}^{{2}}}}$$
$$\displaystyle=\frac{{2}}{{{\left({x}+{3}\right)}^{{2}}{\left({x}-{3}\right)}^{{2}}}}$$
The partial fraction:
$$\displaystyle=\frac{{A}}{{{x}+{3}}}+\frac{{B}}{{{\left({x}+{3}\right)}^{{2}}}}+\frac{{C}}{{{x}-{3}}}+\frac{{D}}{{{\left({x}-{3}\right)}^{{2}}}}$$
Where A-D are constants.