How do you integrate $\int \frac{1}{\sqrt{x} (1 + x)}$ $int 1/(sqrtx(1+x))$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{x} (1 + x)}$ $int 1/(sqrtx(1+x))$ by trigonometric substitution?

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Answer 1 · 2016-09-11T04:20:06

$2 arctan (\sqrt{x}) + C$ $2arctan(sqrtx)+C$

Explanation:

We have the integral:

$\int \frac{d x}{\sqrt{x} (1 + x)}$ $intdx/(sqrtx(1+x))$

We will use the substitution $\sqrt{x} = tan θ$ $sqrtx=tantheta$ . This implies that $x = {tan}^{2} θ$ $x=tan^2theta$ and that $\frac{1}{2 \sqrt{x}} d x = {sec}^{2} θ d θ$ $1/(2sqrtx)dx=sec^2thetad theta$ . Rearranging:

$= 2 \int \frac{d x}{2 \sqrt{x} (1 + x)} = 2 \int \frac{{sec}^{2} θ d θ}{1 + {tan}^{2} θ}$ $=2intdx/(2sqrtx(1+x))=2int(sec^2thetad theta)/(1+tan^2theta)$

Through the Pythagorean identity, $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ .

$= 2 \int \frac{{sec}^{2} θ}{{sec}^{2} θ} d θ = 2 \int d θ = 2 θ + C$ $=2intsec^2theta/sec^2thetad theta=2intd theta=2theta+C$

From $\sqrt{x} = tan θ$ $sqrtx=tantheta$ we see that $θ = arctan (\sqrt{x})$ $theta=arctan(sqrtx)$ . Thus:

$= 2 arctan (\sqrt{x}) + C$ $=2arctan(sqrtx)+C$

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How do you integrate $\int \frac{1}{\sqrt{x} (1 + x)}$ $int 1/(sqrtx(1+x))$ by trigonometric substitution?

1 Answer

Explanation:

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How do you integrate ∫1√x(1+x)int 1/(sqrtx(1+x)) by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{x} (1 + x)}$ $int 1/(sqrtx(1+x))$ by trigonometric substitution?