How do you find the Integral of $\frac{d x}{\sqrt{x^{2} + 16}}$ $dx/sqrt(x^2+16)$ ?

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How do you find the Integral of $\frac{d x}{\sqrt{x^{2} + 16}}$ $dx/sqrt(x^2+16)$ ?

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Answer 1 · 2015-03-30T17:04:07

$\int \frac{d x}{\sqrt{x^{2} + 16}} d x = \int \frac{1}{4 \sqrt{{(\frac{x}{4})}^{2} + 1}} d x$ $int (dx)/ sqrt(x^2+16) dx = int 1/(4sqrt((x/4)^2+1)) dx$

$= \int \frac{1}{\sqrt{{(\frac{x}{4})}^{2} + 1}} (\frac{1}{4}) d x$ $=int 1/(sqrt((x/4)^2+1))( 1/4)dx$

$u = \frac{x}{4}$ $u = x/4$ , $d u = \frac{1}{4} d x$ $du=1/4 dx$ and $\int \frac{1}{\sqrt{u^{2} + 1}} d u = {sinh}^{- 1} u + C$ $int 1/sqrt(u^2+1) du=sinh^(-1) u +C$

So
$\int \frac{d x}{\sqrt{x^{2} + 16}} d x = {sinh}^{- 1} (\frac{x}{4}) + C$ $int (dx)/ sqrt(x^2+16) dx = sinh^(-1) (x/4) +C$

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How do you find the Integral of $\frac{d x}{\sqrt{x^{2} + 16}}$ $dx/sqrt(x^2+16)$ ?

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How do you find the Integral of $\frac{d x}{\sqrt{x^{2} + 16}}$ $dx/sqrt(x^2+16)$ ?