# Calculate the integral $\int_0^1 \frac{dx}{e^x-e^{-2x}+2}$

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 $$2$$?

• -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 – Prakhar Nagpal Nov 10 '19 at 19:05
• Rather than $\displaystyle \int_0^1 \frac u {u^3+2u^2 - 1} \, du,$ you need $\displaystyle \int_1^e \frac u {u^3+2u^2 - 1} \, du. \qquad$ – Michael Hardy Nov 10 '19 at 19:13

## 3 Answers

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.

• Synthetic division by $u+1$ yields instantly the second factor. – Bernard Nov 10 '19 at 19:15

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

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:

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