generals336

2021-01-15

Evaluate the integral.

$\int (\frac{2}{{x}^{3}}+\frac{1}{\sqrt{x}})dx$

Elberte

Skilled2021-01-16Added 95 answers

Given That

$\int (\frac{2}{{x}^{3}}+\frac{1}{\sqrt{x}})dx$

$=2\left[\frac{{x}^{-3+1}}{-3+1}\right]+\left[\frac{{x}^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}\right]+c[\because \int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c]$

$=-\frac{2}{2}{x}^{-2}+2\sqrt{x}+c$

$=-\frac{1}{{x}^{2}}+2\sqrt{x}+c$

$\therefore \int \left(\frac{2}{{x}^{3}+\frac{1}{\sqrt{x}}}\right)dx=-\frac{1}{{x}^{2}}+2\sqrt{x}+c$

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