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Question:

Calculate the integral

$$\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$$

Attempted solution:

I initially had two approaches. First was recognizing that the denominator looks like a quadratic equation. Perhaps we can factor it.

$$\int_0^1 \frac{dx}{e^x-e^{-2x}+2} = \int_0^1 \frac{dx}{e^{-2x}(e^x+1)(e^x+e^2x-1)}$$

To me, this does not appear productive. I also tried factoring out $e^x$ with a similar unproductive result.

The second was trying to make it into a partial fraction. To get to a place where this can efficiently be done, I need to do a variable substitution:

$$\int_0^1 \frac{dx}{e^x-e^{-2x}+2} = \Big[ u = e^x; du = e^x \, dx\Big] = \int_1^e \frac{u}{u^3+2u^2 - 1} \, du$$

This looks like partial fractions might work. However, the question is from a single variable calculus book and the only partial fraction cases that are covered are denominators of the types $(x+a), (x+a)^n, (ax^2+bx +c), (ax^2+bx +c)^n$, but polynomials with a power of 3 is not covered at all. Thus, it appears to be a "too difficult" approach.

A third approach might be to factor the new denominator before doing partial fractions:

$$\int_1^e \frac{u}{u^3+2u^2 - 1} \, du = \int_1^e \frac{u}{u(u^2+2u - \frac{1}{u})} \, du$$

However, even this third approach does not have a denominator that is suitable or partial fractions, since it lacks a u-free term.

What are some productive approaches that can get me to the end without restoring to partial fractions from variables with a power higher than $2$?

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    $\begingroup$ -1 is a root of the bottom expression you can split it and write it as $\int_0 ^1 \frac{u}{(u+1)(pu^2+qu +r)}$ and then use partial fractions $\endgroup$ – Prakhar Nagpal Nov 10 '19 at 19:05
  • $\begingroup$ Rather than $\displaystyle \int_0^1 \frac u {u^3+2u^2 - 1} \, du,$ you need $\displaystyle \int_1^e \frac u {u^3+2u^2 - 1} \, du. \qquad$ $\endgroup$ – Michael Hardy Nov 10 '19 at 19:13
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hint

If you put $u=e^x $, the integral becomes

$$\int_1^e\frac{u\,du}{u^3-1+2u^2}$$ but

$$u^3+2u^2-1=(u+1)(u^2+au+b)$$ with $$1+a=2$$ $$b=-1$$ hence $$u^3+2u^2-1=(u+1)(u^2+u-1)$$ $$=(u+1)(u-\frac{-1-\sqrt{5}}{2})(u-\frac{-1+\sqrt{5}}{2})$$

Now use partial fraction decomposition.

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  • $\begingroup$ Synthetic division by $u+1$ yields instantly the second factor. $\endgroup$ – Bernard Nov 10 '19 at 19:15
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Since $-1$ is a root of $u^3+2u^2-1$, you can write it as $u+1$ times a quadratic monic polynomial. It turns out that that polynomial is $u^2+u-1$. Besides$$\frac u{u^3+2u^2-1}=\frac1{u+1}+\frac{-u+1}{u^2+u-1}.$$Can you take it from here?


Note: There is an error in your computations: the integral that you should be computing is$$\int_1^e\frac u{u^3+2u^2-1}\,\mathrm du.$$

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The change in $u=e^x$

leads to a denominator of degree $3$:

$\displaystyle\int\dfrac{u}{u^3+2u^2-1}\mathop{du}=-\frac 12\ln\big|u^2+u-1\big|-\frac{3\sqrt{5}}{5}\tanh^{-1}\left(\frac{\sqrt{5}}5(2u+1)\right)+\ln\big|u+1\big|+C$


It is possible to do slightly better, while considering $\tanh$.

Since this function is symmetrical in $\pm x$, we take the middle point from $e^x,e^{-2x}$ which is $e^{-x/2}$.


The change $u=\tanh(-x/2)$

leads to a denominator of degree only $2$ which is simpler:

$\displaystyle\int-\dfrac{u-1}{u^2+4u-1}\mathop{du}=-\frac 12\ln\big|u^2+4u-1\big|-\frac {3\sqrt{5}}5\tanh^{-1}\left(\frac{\sqrt{5}}5(u+2)\right)+C$

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