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

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Answer 1 · 2018-05-22T08:47:01

The answer is $= \sqrt{1 + x^{2}} + C$ $=sqrt(1+x^2)+C$

Let $x = tan u$ $x=tanu$

$\Rightarrow$ $=>$ , $d x = {sec}^{2} u d u$ $dx=sec^2udu$

$\sqrt{x^{2} + 1} = \sqrt{1 + {tan}^{2} u} = sec u$ $sqrt(x^2+1)=sqrt(1+tan^2u)=secu$

The integral is

$I = \int \frac{x d x}{\sqrt{x^{2} + 1}} = \int \frac{tan u {sec}^{2} u d u}{sec u}$ $I=int(xdx)/sqrt(x^2+1)=int(tanusec^2udu)/(secu)$

$= \int tan u sec u d u$ $=inttanusecudu$

The derivative of $sec u$ $secu$ is

$=(1/cosu)'=-1/cos^2u*-sinu=tanusecu$

Therefore,

The integral is

$I=secu=sec(arctanx)=sqrt(1+x^2)+C$

How do you integrate int x/sqrt(x^2+1)∫x√x2+1int x/sqrt(x^2+1) by trigonometric substitution?