how you incorporate \$frac1sqrt1+x^2\$ using following substitution? \$1+x^2=t\$ \$Rightarrow\$ \$x=sqrtt-1 Rightarrow dx = fracdt2sqrtt-1dt\$... Now I"m stuck. Ns don"t know how to proceed using substitution rule. By a substitution you suggested you get\$\$int frac12sqrtt(t-1) ,dt=int frac1sqrt4t^2-4t ,dt=int frac1sqrt(2t-1)^2-1 ,dt\$\$Now ns substitution \$u=2t-1\$ seems reasonable.

Você está assistindo: Integral de 1/raiz de x

However her original integral can additionally be addressed by\$x=sinh t\$ e \$dx=cosh t, dt\$ i beg your pardon gives\$\$int fraccosh tcosh t , dt = int 1, dt=t=operatornamearcsinh x = ln (x+sqrtx^2+1)+C,\$\$since \$sqrt1+x^2=sqrt1+sinh^2 t=cosh t\$.

See hyperbolic functions and their inverses.

If you estão familiar (=used to manipulate) with the hyperbolic features then \$x=asinh t\$ is worth trying whenever friend see the expression \$sqrta^2+x^2\$ in your integral (\$a\$ being an arbitrarily constant).

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edited Sep 23 in ~ 7:00 thomas Andrews
answered Aug 5 "12 in ~ 14:00 martinho SleziakMartin Sleziak
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\$egingroup\$ How a partir de you get from \$int frac1sqrt1+x^2 dx\$ to \$int frac1cosh tdx=int fraccosh tcosh tdt\$? \$endgroup\$
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Aug 5 "12 at 14:27

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A different of a hyperbolic role substitution is come let \$x=frac12left(t-frac1t ight)\$. Keep in mind that \$1+x^2=frac14left(t^2+2+frac1t^2 ight)\$.

So \$sqrt1+x^2=frac12left(t+frac1t ight)\$. That was the whole point of a substitution, it is naquela rationalizing substitution that makes the quadrado root simple. Also, \$dx=frac12left(1+frac1t^2 ight),dt\$.

Carry fora the substitution. "Miraculously," our integrante simplifies come \$int fracdtt\$.

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reply Aug 5 "12 at 15:26 andré NicolasAndré Nicolas
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Put \$x= an y\$, so that \$dx=sec^2y dy\$ e \$sqrt1+x^2=sec y\$

\$\$int frac1sqrt1+x^2 dx\$\$

\$\$= int fracsec^2y dysec y\$\$

\$\$=int sec y, dy\$\$

which evaluate to \$displaystyleln|sec y+ an y|+ C\$ , applying the standard formula who proof is here e \$C\$ is an indeterminate constant ao any unknown integral.

\$\$=ln|sqrt1+x^2+x| + C\$\$

We deserve to substitute \$x\$ with \$a sec y\$ para \$sqrtx^2-a^2\$, and with \$a sin y\$ para \$sqrta^2-x^2\$

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edited Aug 5 "12 in ~ 14:37
reply Aug 5 "12 in ~ 14:05 lab bhattacharjeelab bhattacharjee
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\$\$A=intfrac1sqrt<>1+x^2\$\$

Let, \$x = an heta\$

substitute, \$x\$, \$dx\$

\$\$A=intsec hetaleft(fracsec heta + an hetasec heta + an heta ight)d heta\$\$

\$\$A=intleft(fracsec^2 heta + sec heta an hetasec heta + an heta ight)d heta\$\$

Let, \$(sec heta + an heta) = u\$

\$(sec^2 heta + sec heta an heta)d heta = du\$

\$\$A=intfracduu\$\$

\$\$A=lnu+c\$\$

\$\$A=lnvertsec heta + an hetavert+c\$\$

\$\$A=lnvertsqrt<>1+ an^2 heta + an hetavert+c\$\$

\$A=lnvertsqrt<>1+x^2 + xvert+c\$, wherein \$c\$ is a constant

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reply Aug 5 "12 at 17:37
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