How do you integrate $\int [x \sqrt{x^{2} + 1}] d x$ $int [xsqrt(x^2 + 1)] dx$ ?

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How do you integrate $\int [x \sqrt{x^{2} + 1}] d x$ $int [xsqrt(x^2 + 1)] dx$ ?

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Answer 1 · 2015-06-08T00:47:41

This is of the form:

$\sqrt{x^{2} + a^{2}}$ $sqrt(x^2 + a^2)$

which means you can use the substitution:

$x = a tan θ$ $x = atantheta$

So, let:
$x = tan θ$ $x = tantheta$
$d x = {sec}^{2} θ d θ$ $dx = sec^2thetad theta$
$\sqrt{x^{2} + 1} = \sqrt{{tan}^{2} θ + 1} = \sqrt{{sec}^{2} θ} = sec θ$ $sqrt(x^2+1) = sqrt(tan^2theta+1) = sqrt(sec^2theta) = sectheta$

$\int x \sqrt{x^{2} + 1} d x = \int tan θ sec θ {sec}^{2} θ d θ$ $int xsqrt(x^2+1)dx = int tanthetasecthetasec^2thetad theta$

$= \int ({sec}^{2} θ) (sec θ tan θ) d θ$ $= int (sec^2theta)(secthetatantheta)d theta$

Notice how you can do a second substitution here, with u-substitution.

Let:
$u = sec θ$ $u = sectheta$
$d u = sec θ tan θ d θ$ $du = secthetatanthetad theta$

$= \int u^{2} d u = \frac{u^{3}}{3} + C$ $= int u^2du = u^3/3 + C$

$= \frac{{sec}^{3} θ}{3} + C$ $= sec^3theta/3 + C$

Since $sec θ = \sqrt{x^{2} + 1}$ $sectheta = sqrt(x^2 + 1)$ :

$\Rightarrow \frac{1}{3} {(x^{2} + 1)}^{\frac{3}{2}} + C$ $=> 1/3(x^2+1)^(3/2) + C$

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