Integral of arsinhx using logarithmic form?

Question

I have came across a question which has already took a lot of my time, basically it states that I must use the relationship that arsinhx = ln (x+sqrt (x×x+1))
to integrate arsinhx.
I already know how to integrate arsinhx, but I couldn't really manage to integrate it in logarithmic form.
I tried using by parts and substitutions, but it quickly got messy and very tedious, is there a clean way to achieve this?
Thanks in advance!

cidyah · Accepted Answer

∫ arcsinh x dx = ∫  ln ( x+ sqrt(x^2+1) ) dx

Integrate by parts

dv= dx; v= x

u = ln ( x + sqrt(1+x^2) )
du = 1 / ( x + sqrt(1+x^2) ) d/dx ( x + sqrt(1+x^2) )
du = 1 / ( x + sqrt(1+x^2) ) ( 1 + 2x / 2sqrt(1+x^2) )
du = ( 1+ x / sqrt(1+x^2) ) / ( x + sqrt(1+x^2) )
du = ( sqrt(1+x^2) + x ) / (sqrt(1+x^2) ( x+sqrt(1+x^2))
du = 1 / sqrt(1+x^2) dx

∫ u dv = u v - ∫ v du
∫  ln ( x+ sqrt(x^2+1) ) dx = x ln ( x + sqrt(1+x^2) ) -  ∫ x dx / sqrt(1+x^2) ----------- (1)

Consider the integral ∫ x dx / sqrt(1+x^2)

Let u=1+x^2
du = 2x dx
x dx = (1/2) du
∫ x dx / sqrt(1+x^2) = (1/2) ∫ du/ sqrt(u) = sqrt(u) = sqrt( 1+x^2)

substitute this into (1)
∫  ln ( x+ sqrt(x^2+1) ) dx = x ln ( x + sqrt(1+x^2) ) - sqrt (1+x^2) + C

Ian H · Answer

Use relationship that arcsinh(x) = ln[x + √(x^2 + 1)] to integrate arcsinh(x)

Let 2x = e^y – e^(-y) = 2sinh(y)    …………………………(1)
so that y = arcsinh(x) = sinh^-1(x)
(e^y)^2 – 2xe^y – 1 = 0
e^y = x ± √(x^2 + 1), with e^y > 0,  so that explains why 
ln(e^y) = y = sinh^-1(x) = ln[x + √(x^2 + 1)]  …… (2)

d/dx[ln(x)*x] = ln(x) + 1,   (using the product rule) 
and from that we can work out  ∫ln(x)dx = xln(x) – x
We can use that approach to find the integral of (2) by asuming that
it will contain the term xln[x + √(x^2 + 1)]

d/dx{ ln[x + √(x^2 + 1)] * x} = ln[x + √(x^2 + 1)] + T, where
T = x*d/dx{ ln[x + √(x^2 + 1)]}
T = x/[x + √(x^2 + 1)] * [1 + x/√(x^2 + 1)] which rearranges as
T = x/[x + √(x^2 + 1)] * [x + √(x^2 + 1)] /√(x^2 + 1)
That looks messy until you notice that the squared bracket terms cancel.
T = x/√(x^2 + 1)]      This tells us that 
∫ln[x + √(x^2 + 1)]dx = xln[x + √(x^2 + 1)] - ∫[x/√(x^2 + 1)]dx     
Substituting back and integrating the last term
∫arcsinh(x)dx = x*arcsinh(x) - √(x^2 + 1)

Regards – Ian H

Nick · Answer

This will be useful:

d/dx(e^(arcsinhx)) = e^(arcsinhx)*d/dx(arcsinhx) = 1+x/(x^2+1)^(1/2) = e^(arcsinhx)/(x^2+1)^(1/2)

=> d/dx(arcsinhx) = 1/(x^2+1)^(1/2)

Now integrate by parts:

int(arcsinhx dx) = xarcsinhx - int(x/(x^2+1)^(1/2) dx)

= xarcsinhx - (x^2+1)^(1/2) + constant <----

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Integral of arsinhx using logarithmic form?

3 Answers