Find the following integral. \int(\frac{2}{x^{3}}+\frac{1}{\sqrt{x}})dx

Question
Applications of integrals
asked 2020-12-13
Find the following integral.
\(\displaystyle\int{\left({\frac{{{2}}}{{{x}^{{{3}}}}}}+{\frac{{{1}}}{{\sqrt{{{x}}}}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2020-12-14
Given that
\(\displaystyle\int{\left({\frac{{{2}}}{{{x}^{{{3}}}}}}+{\frac{{{1}}}{{\sqrt{{{x}}}}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\left[{\frac{{{x}^{{{3}+{1}}}}}{{-{3}+{1}}}}\right]}+{\left[{\frac{{{x}^{{{\frac{{-{1}}}{{{2}}}}+{1}}}}}{{{\frac{{-{1}}}{{{2}}}}+{1}}}}\right]}+{C}\ {\left[\because\int{x}^{{{n}}}{\left.{d}{x}\right.}={\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}}+{C}\right]}\)
\(\displaystyle={\frac{{-{2}}}{{{2}}}}{x}^{{-{2}}}+{2}\sqrt{{{x}}}+{C}\)
\(\displaystyle=-{\frac{{{1}}}{{{x}^{{{2}}}}}}+{2}\sqrt{{{x}}}+{C}\)
\(\displaystyle\therefore\int{\left({\frac{{{2}}}{{{x}^{{{3}}}}}}+{\frac{{{1}}}{{\sqrt{{{x}}}}}}\right)}{\left.{d}{x}\right.}=-{\frac{{{1}}}{{{x}^{{{2}}}}}}+{2}\sqrt{{{x}}}+{C}\)
0

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