I don"t know how to solve the very first right hand side integral. I tried to let \$u^2 - 1 = t\$, so it becomes

\$\$intfrac1sqrtu^2 - 1=intfrac1sqrttfrac12u , dt = int frac1sqrtt frac12 frac1sqrtt-1 , dt,\$\$

which doesn"t seem to work. The correct answer appears to it is in

\$\$ln left(sqrtx^2 - 1 - x ight)\$\$

But just how do we get this?

You are watching: Integral of 1/sqrt(x^2+1)

integration ordinary-differential-equations
re-superstructure
point out
monitor
edited Sep 8 "16 in ~ 23:28

CommunityBot
1
asked Sep 8 "16 in ~ 22:16

3x89g23x89g2
\$endgroup\$
include a comment |

8
\$egingroup\$
\$\$int frac dx sqrt x^ 2 -1 =int frac sinh t sinh t dt=t +C\ \ x=cosh t \ dx=sinh t \ x=frac e ^ t + e ^ -t 2 \ 2x e ^ t = e ^ 2t +1\ e ^ 2t -2x e ^ t +1=0\ e ^ t =x+sqrt x ^ 2 -1 \$\$

\$\$t=ln \$\$

share
mention
monitor
answer Sep 8 "16 in ~ 22:27

haqnaturalhaqnatural
\$endgroup\$
1
\$egingroup\$
LEt \$u=cosh t\$. You may need to uncover a formula because that arccosh x

re-publishing
cite
follow
answer Sep 8 "16 at 22:20

Empy2Empy2
\$endgroup\$
1
\$egingroup\$
Let \$x=sec t, ; dx=sec t an t,dt\$ come get

\$displaystyleintfrac1sqrtx^2-1dx=intfracsec t an t an tdt=intsec t, dt=lnig|sec t+ an tig|+C=lnig|x+sqrtx^2-1ig|+C\$

re-publishing
point out
monitor
answer Sep 8 "16 at 23:06

user84413user84413
\$endgroup\$
include a comment |
0

See more: Welcome To Sugar Loaf Senior Living In Winona, Mn, Sugar Loaf Senior Living

\$egingroup\$
If the hyperbolic trigonometry is not known, we have the right to see that

\$intfracdtsqrttsqrtt+1=intfracsqrtt+1sqrtt-fracsqrt tsqrtt+1dt.\$

Then \$z=1+frac1t\$ gives \$t=frac1z-1\$, \$dt=frac-dz(1-z)^2\$ and

\$frac12intfracsqrtt+1sqrtt-fracsqrt tsqrtt+1dt=frac12intfrac(1-z)dzsqrtz(1-z)^2=frac12intfracdzsqrtz(1-z)=frac14intfrac1sqrt z(1-sqrtz)+frac1sqrt z(1+sqrtz)dz.\$

It comes

\$intfracdusqrtu^2-1=frac12left(ln(1+sqrt z)-ln(sqrt z-1) ight)=frac12lnleft(frac1+sqrt zsqrt z-1 ight)=frac12lnleft(fracsqrt t+sqrt1+tsqrt1+t-sqrt t ight)=frac12lnleft(fracu+sqrtu^2-1u-sqrtu^2-1 ight)\$

and as \$lnleft(frac1u-sqrtu^2-1 ight)=ln(u+sqrtu^2-1)\$, we attain \$intfracdusqrtu^2-1=ln(u+sqrtu^2-1)\$.

We can likewise note the \$intfracdusqrtu^2-1=frac12intfracsqrtu+1sqrtu-1-fracsqrtu-1sqrt1+udu\$ and take the substitution \$t=fracu+1u-1\$.