Find the indefinite integral.
∫((e^-x-1)/(e^-x + x)^2) dx
I am trying to use the substitution method.
u=e^-x + x
du= -e^-x+1 dx
Right so far?
got it figured out.
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Indefinite integral:Show steps
Yes, you are correct so far. Using the substitution method, let u = e^(-x) + x. Then, differentiate u with respect to x to find du/dx.
du/dx = d/dx(e^(-x) + x)
To differentiate e^(-x), we can use the chain rule. The derivative of e^(-x) with respect to x is -e^(-x). The derivative of x with respect to x is 1.
du/dx = -e^(-x) + 1
Now, solve this equation for dx.
du = -e^(-x) + 1 dx
To isolate dx, multiply both sides of the equation by -1.
-dx = e^(-x) - 1 du
Now, substitute the values of dx and du into the integral.
∫((e^(-x) - 1)/(e^(-x) + x)^2) dx = ∫(1/(u^2)) du = ∫u^(-2) du
The indefinite integral of u^(-2) with respect to u is (-1/u) + C, where C is the constant of integration.
Therefore, the indefinite integral of the given function is (-1/(e^(-x) + x)) + C, where C is the constant of integration.