How do you find the antiderivative of $\int \frac{1}{\sqrt{1 + x^{2}}} d x$ $int 1/sqrt(1+x^2) dx$ ?

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Answer 1 · 2017-06-23T04:01:37

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

$I = \int \frac{1}{\sqrt{1 + x^{2}}} d x$ $I=int1/sqrt(1+x^2)dx$

Let $x = tan θ$ $x=tantheta$ . This implies that $d x = {sec}^{2} θ d θ$ $dx=sec^2thetad theta$ .

$I = \int \frac{1}{\sqrt{1 + {tan}^{2} θ}} {sec}^{2} θ d θ$ $I=int1/sqrt(1+tan^2theta)sec^2thetad theta$

Since $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ :

$I = \int sec θ d θ = ln | sec θ + tan θ |$ $I=intsecthetad theta=lnabs(sectheta+tantheta)$

Note that $tan θ = x$ $tantheta=x$ and $sec θ = \sqrt{1 + {tan}^{2} θ} = \sqrt{1 + x^{2}}$ $sectheta=sqrt(1+tan^2theta)=sqrt(1+x^2)$ :

$I = ln ∣ ∣ x + \sqrt{1 + x^{2}} ∣ ∣ + C$ $I=lnabs(x+sqrt(1+x^2))+C$

How do you find the antiderivative of ∫1√1+x2dxint 1/sqrt(1+x^2) dx?