Evaluate the integral. int(2/x^3+1/sqrt x)dx

Evaluate the integral. int(2/x^3+1/sqrt x)dx

Question
Applications of integrals
asked 2021-01-15
Evaluate the integral.
\(\displaystyle\int{\left(\frac{{2}}{{x}^{{3}}}+\frac{{1}}{\sqrt{{x}}}\right)}{\left.{d}{x}\right.}\)

Answers (1)

2021-01-16
Given That
\(\displaystyle\int{\left(\frac{{2}}{{x}^{{3}}}+\frac{{1}}{\sqrt{{x}}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\left[\frac{{{x}^{{-{3}+{1}}}}}{{-{3}+{1}}}\right]}+{\left[\frac{{{x}^{{-\frac{{1}}{{2}}+{1}}}}}{{-\frac{{1}}{{2}}+{1}}}\right]}+{c}{\left[\because\int{x}^{{n}}{\left.{d}{x}\right.}=\frac{{{x}^{{{n}+{1}}}}}{{{n}+{1}}}+{c}\right]}\)
\(\displaystyle=-\frac{{2}}{{2}}{x}^{{-{{2}}}}+{2}\sqrt{{x}}+{c}\)
\(\displaystyle=-\frac{{1}}{{x}^{{2}}}+{2}\sqrt{{x}}+{c}\)
\(\displaystyle\therefore\int{\left(\frac{{2}}{{{x}^{{3}}+\frac{{1}}{\sqrt{{x}}}}}\right)}{\left.{d}{x}\right.}=-\frac{{1}}{{x}^{{2}}}+{2}\sqrt{{x}}+{c}\)
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