How do you integrate $\int \frac{1}{\sqrt{x^{2} + 2 x}}$ $int 1/sqrt(x^2+2x)$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{\sqrt{x^{2} + 2 x}}$ $int 1/sqrt(x^2+2x)$ by trigonometric substitution?

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Answer 1 · 2017-02-20T22:10:21

$ln ∣ ∣ ∣ \sqrt{{(x + 1)}^{2} - 1} + x + 1 ∣ ∣ ∣ + C$ $ln|sqrt((x + 1)^2 - 1) + x + 1| + C$

Explanation:

Complete the square in the denominator (within the $\sqrt{}$ $sqrt$ ).

$\int \frac{1}{\sqrt{1 (x^{2} + 2 x + 1 - 1)}} d x$ $int 1/sqrt(1(x^2 + 2x + 1 - 1))dx$

$\int \frac{1}{\sqrt{1 (x^{2} + 2 x + 1) - 1}} d x$ $int 1/sqrt(1(x^2 + 2x + 1) - 1)dx$

$\int \frac{1}{\sqrt{{(x + 1)}^{2} - 1}} d x$ $int 1/sqrt((x + 1)^2 - 1)dx$

Let $u = x + 1$ $u = x + 1$ . Then $d u = d x$ $du = dx$ .

$\int \frac{1}{\sqrt{u^{2} - 1}} d u$ $int 1/sqrt(u^2 - 1)du$

Now use the substitution $u = sec θ$ $u = sectheta$ . Then $d u = sec θ tan θ d θ$ $du = secthetatanthetad theta$ .

$\int \frac{1}{\sqrt{{sec}^{2} θ - 1}} sec θ tan θ d θ$ $int 1/sqrt(sec^2theta - 1) secthetatantheta d theta$

$\int \frac{1}{\sqrt{{tan}^{2} θ}} sec θ tan θ d θ$ $int 1/sqrt(tan^2theta) secthetatantheta d theta$

$\int \frac{1}{tan θ} sec θ tan θ d θ$ $int 1/tantheta secthetatantheta d theta$

$\int sec θ d θ$ $int sectheta d theta$

This is a known integral that can be derived here

$ln | sec θ + tan θ | + C$ $ln|sectheta + tantheta| + C$

Obviously it's not good enough to stay in $θ$ $theta$ ; we have to return to $x$ $x$ . From our initial substitution, $\frac{u}{1} = sec θ$ $u/1 = sectheta$ . This means that the side opposite $θ$ $theta$ measures $\sqrt{u^{2} - 1}$ $sqrt(u^2 - 1)$ . This also means that $tan θ = \frac{\sqrt{u^{2} - 1}}{1} = \sqrt{u^{2} - 1}$ $tantheta = sqrt(u^2 - 1)/1 = sqrt(u^2 - 1)$ .

$ln ∣ ∣ \sqrt{u^{2} - 1} + u ∣ ∣ + C$ $ln|sqrt(u^2 - 1) + u| + C$

We have one more substitution to reverse.

$ln ∣ ∣ ∣ \sqrt{{(x + 1)}^{2} - 1} + x + 1 ∣ ∣ ∣ + C$ $ln|sqrt((x + 1)^2 - 1) + x + 1| + C$

Hopefully this helps!

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How do you integrate $\int \frac{1}{\sqrt{x^{2} + 2 x}}$ $int 1/sqrt(x^2+2x)$ by trigonometric substitution?

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Explanation:

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