0
$\begingroup$

I am trying to evaluate

$$ \int \frac{\left(2-z\left(u\right)\right)z'\left(u\right)}{\left(z\left(u\right)\right)^2-z\left(u\right)+2}du \quad (1) $$

I think I've found a solution by using integration by substitution but I am having trouble with expressing the infinitesmall. You'll understand as you read.


My attempt

Lets re-write the integral more simply as:

$$\int \frac{(2-z) z'}{z^2-z+2}du$$

Since $z$ is not a variable but a function of $u$, the first basic composite integrals that come to mind are the following:

$$\int \frac{f(x)f'(x)}{f(x)^2+1} dx = \frac12 \ln\left|f^2(x)+1\right| +C \quad (2)$$ $$\int \frac{f'(x)}{f(x)^2+1} dx = \arctan\left|f(x)\right| +C\quad (3)$$

We can re-write the denominator of $(1)$ as

$$z^2-z-2 = \frac74\left[ \left( \frac{2}{\sqrt 7} (z-\frac12) \right)^2 +1 \right]$$

Hence. plugging it in $(1)$

$$\int \frac{(2-z) z'}{z^2-z+2}du = \int \frac{(2-z) z'}{\frac74\left[ \left( \frac{2}{\sqrt 7} (z-\frac12) \right)^2 +1 \right]}du = \frac47 \int \frac{(2-z) z'}{ \left( \frac{2}{\sqrt 7} (z-\frac12) \right)^2 +1 }du$$

It seems we are on the right way. Let's break the integral:

$$ \frac87 \int\frac{z'}{ \left( \frac{2}{\sqrt 7} (z-\frac12) \right)^2 +1 }du - \frac47 \int\frac{zz'}{ \left( \frac{2}{\sqrt 7} (z-\frac12) \right)^2 +1 }du$$

If we could bring the integral in $(2) \text{ and } (3)$ forms then we would have had evaluated it, i.e:

Let $h = \frac{2}{\sqrt 7}(z(u) - \frac12) \to $ $z(u) = \frac{\sqrt 7}{2}h+\frac12$

All we need now is to find $du$, but that I am having trouble understanding how infinitesmall will change by this substitution. Note that $z$ is a function of $u$ and not a variable.

How can $du$ be expressed in terms of $h$?

I think we can write is as:

$$h = \frac{2}{\sqrt 7}(z(u) - \frac12) \iff dh = \frac{2}{\sqrt 7} z'(u) du $$

but I am unsure if this is correct.

$\endgroup$
1
$\begingroup$

I would do so:

$$ \small \begin{aligned} \int \frac{2-z(u)}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u & = \frac{3}{2}\int \frac{1}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u - \frac{1}{2}\int \frac{2\,z(u)-1}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u \\ & = \frac{3}{2}\int \frac{1}{\left(z(u)-\frac{1}{2}\right)^2+\frac{7}{4}}\,z'(u)\,\text{d}u - \frac{1}{2}\int \frac{2\,z(u)-1}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u \\ & = \frac{6}{7}\int \frac{1}{1+\left[\frac{2}{\sqrt{7}}\left(z(u)-\frac{1}{2}\right)\right]^2}\,z'(u)\,\text{d}u - \frac{1}{2}\int \frac{2\,z(u)-1}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u \\ & = \frac{3\sqrt{7}}{7}\int \frac{\frac{2}{\sqrt{7}}}{1+\left[\frac{2}{\sqrt{7}}\left(z(u)-\frac{1}{2}\right)\right]^2}\,z'(u)\,\text{d}u - \frac{1}{2}\int \frac{2\,z(u)-1}{z^2(u)-z(u)+2}\,z'(u)\,\text{d}u \\ & = \frac{3\sqrt{7}}{7}\int \frac{1}{1+h^2(u)}\,h'(u)\,\text{d}u - \frac{1}{2}\int \frac{1}{k(u)}\,k'(u)\,\text{d}u \\ & = \frac{3\sqrt{7}}{7}\,\arctan(h(u)) - \frac{1}{2}\,\ln|k(u)| + c \\ & = \frac{3\sqrt{7}}{7}\,\arctan\left(\frac{2\,z(u)-1}{\sqrt{7}}\right) - \frac{1}{2}\,\ln\left(z^2(u)-z(u)+2\right) + c \,. \end{aligned} $$

$\endgroup$
1
  • 1
    $\begingroup$ Nice one, easy and clear $\endgroup$ – Dimitris Jul 25 '20 at 11:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.