No closed variety answer exists, yet you are able to nonetheless get a chain answer as follows: replace variables via u = ln x. Then exp^u = x, so exp^u du = dx. So thie new essential in terms of u is of (exp^u)/u. Write exp^u as a chain, divid each and every term via u and combine. You get ln u + u + (u^2)/[2*2!] + ... +( u^n)/[n*n!] +.... replace u with ln x to get the suited answer.
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primero, sea u = Ln(x+√(1+x^2)) ; dv = dx
entonces: du = 1/(√(x^2+1)) dx; v = x
por integracion por partes:
integral(udv) = uv - integral(vdu)
Integral(Ln(x+√(1+x^2)) dx)=
(Ln(x+√(1+x^2)))x - integral(x/(√(x^2+1)) dx)
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para resolver integral(x/(√(x^2+1)) dx)
sea r=x^2+1
dr=2xdx
dr/2=xdx
integral(x/(√(x^2+1)) dx)
integral(1/√r dr/2)=√r=√(x^2+1)
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entonces:
Integral(Ln(x+√(1+x^2)) dx)=
(Ln(x+√1+x^2)))x - √(x^2+1)
No closed variety answer exists, yet you are able to nonetheless get a chain answer as follows: replace variables via u = ln x. Then exp^u = x, so exp^u du = dx. So thie new essential in terms of u is of (exp^u)/u. Write exp^u as a chain, divid each and every term via u and combine. You get ln u + u + (u^2)/[2*2!] + ... +( u^n)/[n*n!] +.... replace u with ln x to get the suited answer.
int(ln(x+sqrt(1+x^2)), x) = ln(x + sqrt(1 + x^2))*x - sqrt(1 + x^2) + C