Transforming to 2d input from a derative

Thought always that the round d is reserved for function of two variables x,y , but that seems to be not the case here ?

restart;

Comparing Different Answers

Een antwoord ergens gegeven is

Int(sqrt(x^2+1), x) = (1/2)*x*sqrt(x^2+1)+(1/2)*ln(x+sqrt(x^2+1)) + C (vb)

Mple geeft

>	Int(sqrt(x^2+1),x)=int(sqrt(x^2+1),x)+C[1];

Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*arcsinh(x)+C[1]

(1)

De twee antwoorden lijken nog niet opelkaar !
In het gegeven antwoord staat er een ln en in Maple kan een expressie omgezet worden in ln termen

>	convert(%,ln);

Int((x^2+1)^(1/2), x) = (1/2)*x*(x^2+1)^(1/2)+(1/2)*ln(x+(x^2+1)^(1/2))+C[1]

(2)

(2) is hetzelfde (vb)

Dezelfde integraal i sook gegeven als

>	Int(sqrt(x^2+1),x)=((x+sqrt(x^2+1))^2+4*ln(x+sqrt(x^2+1))-(x+sqrt(x^2+1))^(-2))/8+C[2];

Int((x^2+1)^(1/2), x) = (1/8)*(x+(x^2+1)^(1/2))^2+(1/2)*ln(x+(x^2+1)^(1/2))-(1/8)/(x+(x^2+1)^(1/2))^2+C[2]

(3)

Controle

een effectieve manier om twe antwoorden t evergelijken voor hetzelfde probleem is het verschil te berekenen van een vergelijking met de twee integralen

#lhs(%);

#rhs(%%);

>	#diff(lhs(%)-rhs(%)=0,x);

NULL

>	#diff(f,x);

>	diff(lhs(%)-rhs(%)=0,x);

(x^2+1)^(1/2)-(1/4)*(x+(x^2+1)^(1/2))*(1+x/(x^2+1)^(1/2))-(1/2)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))-(1/4)*(1+x/(x^2+1)^(1/2))/(x+(x^2+1)^(1/2))^3 = 0

(4)

>	simplify(%);

0 = 0

(5)

Strange that diff(lhs(%)-rhs(%)=0,x); is translated by 2 d input with round d notation for functions with two variables ?
The two integrals are functions of one variable
diff(f, x)

Download Controleren_dezelfde_antwoord_voo_expressies.mw

Transforming to 2d input from a derative

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