Asked by regin
find indefinite integral of
(e^-x)-(1)/(e^-x+(x)^2 dx
(e^-x)-(1)/(e^-x+(x)^2 dx
Answers
Answered by
Reiny
my interpretation:
(e^-x - 1)/(e^-x+x^2) , you have mis-matched brackets
even the usual reliable Wolfram had difficulties with that one
http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2Bx%5E2%29&random=false
the way you typed it, fixing the missing bracket
e^-x - (1/(e^-x+x^2) )
http://integrals.wolfram.com/index.jsp?expr=e%5E-x+-+%281%2F%28e%5E-x%2Bx%5E2%29+%29&random=false
same result,
that's a tough one
(e^-x - 1)/(e^-x+x^2) , you have mis-matched brackets
even the usual reliable Wolfram had difficulties with that one
http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2Bx%5E2%29&random=false
the way you typed it, fixing the missing bracket
e^-x - (1/(e^-x+x^2) )
http://integrals.wolfram.com/index.jsp?expr=e%5E-x+-+%281%2F%28e%5E-x%2Bx%5E2%29+%29&random=false
same result,
that's a tough one
Answered by
bobpursley
let u= (e^-x+(x))^-2
then du = -2/(e^-x+(x)) * (-e^-x+(1))dx
which changes the integral to
1/2 u du
or u^2
then du = -2/(e^-x+(x)) * (-e^-x+(1))dx
which changes the integral to
1/2 u du
or u^2
Answered by
Reiny
and Bob saw an even third interpretation of your typing
(e^-x - 1)/(e^-x+ x )^2
http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2B+x+%29%5E2&random=false
(e^-x - 1)/(e^-x+ x )^2
http://integrals.wolfram.com/index.jsp?expr=%28e%5E-x+-+1%29%2F%28e%5E-x%2B+x+%29%5E2&random=false
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