Question

What is `int \ ln(x+sqrt(x^2+1)) \ dx` ?

Answer 1

`int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x sinh^(-1) x - sqrt(x^2+1) + C`

`color(white)(int ln(x+sqrt(x^2+1)) dx) = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C`

Explanation:

Let us try a hyperbolic substitution.

Let `x = sinh theta`

Then:

`int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = int color(white)(.)ln(sinh theta + sqrt(sinh^2 theta + 1)) dx/(d theta) d theta`

`color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(sinh theta + cosh theta) cosh theta color(white)(.)d theta`

`color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)ln(e^theta) cosh theta color(white)(.)d theta`

`color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = int color(white)(.)theta cosh theta color(white)(.)d theta`

`color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = theta sinh theta - cosh theta + C`

`color(white)(int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx) = x sinh^(-1) x - sqrt(x^2+1) + C`

If we would prefer not to have an inverse hyperbolic function in the answer, let's find another expression for it:

Let:

`y = sinh^(-1) x`

Then:

`x = sinh y = 1/2(e^y-e^(-y))`

Hence:

`e^y-2x-e^(-y) = 0`

So:

`(e^y)^2-(2x)(e^y)-1 = 0`

So using the quadratic formula:

`e^y = (2x+-sqrt((2x)^2+4))/2`

`color(white)(e^y) = x+-sqrt(x^2+1)`

If `y` is real then `e^y > 0` and we need the `+` sign here.

So:

`e^y = x+sqrt(x^2+1)`

and:

`y = ln(x+sqrt(x^2+1))`

That is:

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

So:

`int color(white)(.)ln(x+sqrt(x^2+1)) color(white)(.)dx = x ln(x+sqrt(x^2+1)) - sqrt(x^2+1) + C`

Answer 2

Note that

`d/dxln(x+sqrt(x^2+1))=1/(x+sqrt(x^2+1))(1+x/sqrt(x^2+1))`

`color(white)(d/dxln(x+sqrt(x^2+1)))=1/sqrt(x^2+1)`

Then we should try integration by parts:

`intln(x+sqrt(x^2+1))dx`

Letting:

`{(u=ln(x+sqrt(x^2+1)),=>,du=1/sqrt(x^2+1)dx),(dv=dx,=>,v=x):}`

So the integral equals:

`=xln(x+sqrt(x^2+1))-intx/sqrt(x^2+1)dx`

`=xln(x+sqrt(x^2+1))-sqrt(x^2+1)+C`

Answer 3

After using `x=tanu` and `dx=(secu)^2*du` transform,

`intLn(x+sqrt(x^2+1))dx` integral became,

`=intLn(secu+tanu)*(secu)^2*du`

`=tanu*Ln(secu+tanu)-inttanu*[(secu)^2+secu*tanu]/(secu+tanu)*du`

`=tanu*Ln(secu+tanu)-inttanu*secu*(secu+tanu)/(secu+tanu)*du`

`=tanu*Ln(secu+tanu)-intsecu*tanu*du`

`=tanu*Ln(secu+tanu)-secu+C`

`=tanu*Ln(secu+tanu)-sqrt((tanu)^2+1)+C`

`=x*Ln(x+sqrt(x^2+1))-sqrt(x^2+1)+C`

Explanation:

1) I used `x=tanu` and `dx=(secu)^2*du` transform.

2) I solved transformed integral with integration by parts method.

3) I used inverse transform.