Question

What is the integral of `sqrt(x^2 + 1)`?

Answer 1

Please, see the answer below:

Answer 2

`=1/2(sinh^(-1)x+xsqrt(1+x^2))+C`

Explanation:

Use `x = sinh u`, giving `dx = cosh u du`,

`cosh^2u+sinh^2u=cosh 2u and sinh 2u = 2 sinh u cosh u`.

Nowm the given integral becomes

`int cosh^2u du`

`=1/2 int (1+ cosh 2u ) du`

`=1/2[u+1/2sinh 2u]+C`

`=1/2(u+sinh u cosh u)+C`

`=1/2(sinh^(-1)x+xsqrt(1+x^2))+C`.

In log form,

`sinh^(-1)x=ln(x+sqrt(x^2+1))`

Answer 3

I got:

`1/2xsqrt(x^2 + 1) + 1/2ln|sqrt(x^2 + 1) + x| + C`

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Here's another way to do it, without using reduction formulas or hyperbolic functions.

Let:

`x = tantheta`
`dx = sec^2thetad theta`

`=> int sqrt(tan^2theta + 1)sec^2thetad theta`

`= int sec^3theta d theta`

`= int sectheta(sec^2theta)d theta`

This can be solved using integration by parts. Let:

`u = sectheta`
`dv = sec^2thetad theta`
`v = tantheta`
`du = secthetatanthetad theta`

`=> uv - intvdu`

`= secthetatantheta - int secthetatan^2thetad theta`

`= secthetatantheta - int sec^3theta - secthetad theta`

`= secthetatantheta - int sec^3thetad theta + intsecthetad theta`

We see the integral reappears. Thus:

`=> 2int sec^3thetad theta = secthetatantheta + intsecthetad theta`

`=> int sec^3thetad theta = 1/2secthetatantheta + 1/2ln|sectheta + tantheta|`

Finally, un-substitute. Since `x = tantheta`, `sqrt(x^2 + 1) = sqrt(tan^2theta + 1) = sqrt(sec^2theta) = sectheta`, and we have:

`=> int sqrt(x^2 + 1)dx = color(blue)(1/2xsqrt(x^2 + 1) + 1/2ln|sqrt(x^2 + 1) + x| + C)`