Asked by Steff
integrate (e^x-x)^2 dx
Answers
Answered by
Reiny
I would expand it first
∫ (e^x - x)^2 dx
= ∫ e^(2x) - 2xe^x + x^2 dx
let's concentrate on the middle term, since the other two terms are easy to integrate
∫ 2xe^x dx
let u = 2x , let dv = e^x dx
du = 2dx and v = e^x
then ∫ 2xe^x dx = (2x)(e^x) - ∫ 2e^x dx
= 2xe^x - 2e^x
finally ...
∫ (e^x-x)^2 dx
= ∫ e^(2x) - 2xe^x + x^2 dx
= (1/2e^(2x) - (2xe^x - 2e^x) + (1/3)x^3 + c
= (1/2)e^(2x) - 2xe^x + 2e^x + (1/3)x^3 + c
∫ (e^x - x)^2 dx
= ∫ e^(2x) - 2xe^x + x^2 dx
let's concentrate on the middle term, since the other two terms are easy to integrate
∫ 2xe^x dx
let u = 2x , let dv = e^x dx
du = 2dx and v = e^x
then ∫ 2xe^x dx = (2x)(e^x) - ∫ 2e^x dx
= 2xe^x - 2e^x
finally ...
∫ (e^x-x)^2 dx
= ∫ e^(2x) - 2xe^x + x^2 dx
= (1/2e^(2x) - (2xe^x - 2e^x) + (1/3)x^3 + c
= (1/2)e^(2x) - 2xe^x + 2e^x + (1/3)x^3 + c
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