Question

evaluate f 9/1. x+1 /√X dx

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Answer 1

$\boxed{\mathrm{\int\dfrac{x+1}{\sqrt{x}}dx=\dfrac{2}{3}x\sqrt{x}+2\sqrt{x}+C}}$

Explanation:

Now, $\mathrm{\dfrac{x+1}{\sqrt{x}}}$

$\mathrm{=\sqrt{x}+\dfrac{1}{\sqrt{x}}}$

$\mathrm{=x^{1/2}+x^{-1/2}}$

Taking integration on both sides, we get

$\mathrm{\int\dfrac{x+1}{\sqrt{x}}dx=\int (x^{1/2}+x^{-1/2})dx}$

$\mathrm{\Rightarrow \int\dfrac{x+1}{\sqrt{x}}dx=\dfrac{1}{1/2+1}x^{1/2+1}+\dfrac{1}{-1/2+1}x^{-1/2+1}+C}$

where $\mathrm{C}$ is constant of integration

$\mathrm{\Rightarrow \int\dfrac{x+1}{\sqrt{x}}dx=\dfrac{1}{3/2}x^{3/2}+\dfrac{1}{1/2}x^{1/2}+C}$

$\mathrm{\Rightarrow \int\dfrac{x+1}{\sqrt{x}}dx=\dfrac{2}{3}x\sqrt{x}+2\sqrt{x}+C}$

This is the required integral.