find indefinite integral of
(e^-x)-(1)/(e^-x+(x)^2 dx
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
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
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
To find the indefinite integral of the given function, let's break it down:
∫[(e^-x) - (1)/(e^-x + (x)^2)] dx
To integrate this expression, we can apply the technique of substitution. Let's introduce a new variable, u, which is equal to e^(-x):
u = e^(-x)
To find du/dx, we differentiate both sides of the equation with respect to x:
du/dx = d/dx(e^(-x))
du/dx = -e^(-x)
We can then solve this equation for dx:
dx = -du/u
Substituting the values into our integral expression:
∫[(e^-x) - (1)/(e^-x + (x)^2)] dx
= ∫[(u) - (1)/(u + (x)^2)] (-du/u)
= -∫[1 - (1)/(u + (x)^2)] du
Now, let's focus on the integral term:
∫[1 - (1)/(u + (x)^2)] du
This can be further simplified as:
∫1 du - ∫(1)/(u + (x)^2) du
The first term, ∫1 du, integrates to u:
u - ∫(1)/(u + (x)^2) du
To find the indefinite integral of (1)/(u + (x)^2), we can use the partial fraction decomposition.
Let A be a constant, then:
(1)/(u + (x)^2) = A/(u + (x)^2) + B(u + (x)^2)
=> 1 = A(u + (x)^2) + B(u + (x)^2)
By equating the coefficients of like terms:
0 = (A + B)u + (A + B)(x)^2
Since u and (x)^2 are independent, the coefficients must be individually equal to zero:
A + B = 0
A + B = 1
Solving these two equations simultaneously, we find:
A = 1/2
B = -1/2
Therefore:
(1)/(u + (x)^2) = (1/2)/(u + (x)^2) - (1/2)(u + (x)^2)
Returning to our integral expression:
∫[1 - (1)/(u + (x)^2)] du
= u - ∫((1/2)/(u + (x)^2) - (1/2)(u + (x)^2)) du
= u - (1/2)ln|u + (x)^2| - (1/2)u + (1/4)(u + (x)^2)^2 + C
Finally, substituting back u = e^(-x):
= e^(-x) - (1/2)ln|e^(-x) + (x)^2| - (1/2)e^(-x) + (1/4)(e^(-x) + (x)^2)^2 + C
Therefore, the indefinite integral of (e^-x) - (1)/(e^-x + (x)^2) dx is:
e^(-x) - (1/2)ln|e^(-x) + (x)^2| - (1/2)e^(-x) + (1/4)(e^(-x) + (x)^2)^2 + C.