How do you integrate $\int \sqrt{x} \sqrt{x \sqrt{x} + 1}$ $int sqrtxsqrt(xsqrtx+1)$ using substitution?

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How do you integrate $\int \sqrt{x} \sqrt{x \sqrt{x} + 1}$ $int sqrtxsqrt(xsqrtx+1)$ using substitution?

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Answer 1 · 2016-09-27T23:06:50

${(\frac{2}{3})}^{2} {(x^{\frac{3}{2}} - 1)}^{\frac{3}{2}} + C$ $(2/3)^2(x^(3/2)-1)^(3/2) + C$

Explanation:

This integral is a power integral. The square root operator helps to conceal de true problem.

$\sqrt{x} \sqrt{x \sqrt{x} + 1} = x^{\frac{1}{2}} {(x^{\frac{3}{2}} - 1)}^{\frac{1}{2}} = {(\frac{2}{3})}^{2} \frac{d}{d x} {(x^{\frac{3}{2}} - 1)}^{\frac{3}{2}}$ $sqrtxsqrt(xsqrtx+1) = x^(1/2)(x^(3/2)-1)^(1/2)=(2/3)^2d/dx(x^(3/2)-1)^(3/2)$

So

$\int \sqrt{x} (\sqrt{x \sqrt{x} + 1}) d x = {(\frac{2}{3})}^{2} {(x^{\frac{3}{2}} - 1)}^{\frac{3}{2}} + C$ $int sqrtx(sqrt(xsqrtx+1))dx = (2/3)^2(x^(3/2)-1)^(3/2) + C$

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How do you integrate $\int \sqrt{x} \sqrt{x \sqrt{x} + 1}$ $int sqrtxsqrt(xsqrtx+1)$ using substitution?

1 Answer

Explanation:

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