How do you integrate (\frac{e^x}{x})dx ?

Ashley Bell 2021-12-13 Answered
How do you integrate (exx)dx ?
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Expert Answer

Laura Worden
Answered 2021-12-14 Author has 45 answers
This is sometimes called the exponential integral:
exxdx=Ei(x)+C
But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for ex
ex=1+x+x22!+x33!+=n=0xnn!
Then:
exx=1x+1+x2!+x23!+=1x+n=0xn(n+1)!
So the antiderivative will be:
exxdx=(1x+1+x2!+x23!+)dx
=ln(|x|)+x+x222!+x333!++C
exxdx=ln(|x|)+n=1xnnn!+C
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Joseph Lewis
Answered 2021-12-15 Author has 43 answers
Answer: ex=x00+x11+x22+x33+
=1+x+x22+x33+
exxdx
=1x(1+x+x22+x33+)dx
=(1x+1+x2+x23+)dx
=log|x|+x+x222+x333
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Why is this solution incorrect?
Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then 1PK2+1QK2 is the same for all positions of the chord.
If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for y2=4ax.
let the point K be (c,0)
Equation of line PQ using parametric coordinates: x=c+rcosθ (Equation 1), y=rsinθ (Equation 2)
From equation 1 and 2: (xc)2+y2=r2 (Equation 3)
Using Equation 3 and y2=4ax, we get this quadratic in r: r24ax(xk)2=0 (Equation 4)
Roots of this quadratic (r1 and r2) would be the lengths of PK and QK
From Equation 4, r1+r2=0 and r1r2=(4ax+(xk)2)
We know, 1PK2+1QK2=1r12+1r22=(r1+r2)22r1r2(r1r2)2=2r1r2
As value of r1r2 is not constant thus 1PK2+1QK2 does not turn out to be constant. Hence, this solution is incorrect.
I've seen the correct solution but I wanted to know why this solution is incorrect?

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