A) \[\frac{{{x}^{5}}}{5}-\frac{2{{x}^{3}}}{3}+2x+C\]
B) \[\frac{{{x}^{5}}}{5}-\frac{2{{x}^{3}}}{3}-2x+C\]
C) \[\frac{{{x}^{5}}}{5}+\frac{2{{x}^{3}}}{3}-2x+C\]
D) \[\frac{{{x}^{5}}}{5}+\frac{2{{x}^{3}}}{3}+2x+C\]
Correct Answer: D
Solution :
We have given that \[\int_{{}}^{{}}{\frac{{{x}^{8}}+4}{{{x}^{4}}-2{{x}^{2}}+2}}dx=\int_{{}}^{{}}{\frac{({{x}^{8}}+4=4{{x}^{4}})-4{{x}^{4}}}{({{x}^{4}}-2{{x}^{2}}+2)}}dx\] \[=\int_{{}}^{{}}{\frac{{{({{x}^{4}}+2)}^{2}}-{{(2{{x}^{2}})}^{2}}}{({{x}^{4}}-2{{x}^{2}}+2)}}dx\] \[=\int_{{}}^{{}}{\frac{({{x}^{4}}+2+2{{x}^{2}})({{x}^{4}}+2-2{{x}^{2}})}{({{x}^{4}}-2{{x}^{2}}+2)}}dx\] \[=\int_{{}}^{{}}{({{x}^{4}}+2{{x}^{2}}+2)}dx\] \[=\frac{{{x}^{5}}}{5}+\frac{2}{3}{{x}^{3}}+2x+C\]
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