-w8.7.28 integral rational expression

Click For Summary

Discussion Overview

The discussion revolves around the integral of a rational expression, specifically the integral $$\int\frac{t+1}{{t}^{2}+t-1}\ dt$$. Participants explore various methods of integration, including partial fraction decomposition and substitutions, while addressing the challenges posed by the non-factorable denominator.

Discussion Character

  • Technical explanation, Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants note that the denominator does not factor with rational roots and suggest using the quadratic formula to find the roots.
  • Others propose expressing the integral in terms of partial fractions, with one participant providing a specific decomposition involving square roots.
  • There is a discussion about integrating the resulting expressions, with participants presenting different approaches and substitutions for the integrals.
  • One participant suggests an alternative method involving a substitution that simplifies the integral into a more manageable form.
  • Several participants express uncertainty about the correctness of their approaches and the results, leading to corrections and refinements of earlier claims.
  • One participant ultimately presents a final expression for the integral that aligns with a provided book answer, but there is no consensus on the methods used to arrive at this result.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for solving the integral, with multiple competing views and approaches remaining throughout the discussion.

Contextual Notes

Some methods proposed depend on specific substitutions or assumptions about the integrand, and there are unresolved mathematical steps in the integration process.

karush
Gold Member
MHB
Messages
3,240
Reaction score
5
nmh{1000}
$\tiny{\text {Whitman 8.7.28 integral rational expression}} $
$$\displaystyle
\int\frac{t+1}{{t}^{2}+t-1}\ dt$$
$\text{book answer}$
$$\displaystyle\frac{5+\sqrt{5}}{10}
\ln\left({2t+1-\sqrt{5}}\right)
+\frac{5-\sqrt{5}}{10}
\ln\left({2t+1+\sqrt{5}}\right)+C$$

$\text{expansion}$
$$\displaystyle
\int\frac{t+1}{{t}^{2}+t-1}\ dt
=\int\frac{t}{{t}^{2}+t-1}\ dt
+\int\frac{1}{{t}^{2}+t-1}\ dt $$

Not sure how to approach this since it won't factor
 
Last edited:
Physics news on Phys.org
Well, the denominator of the integrand doesn't factor with rational roots being the result, but you can get the roots using the quadratic formula and then factor that way...you will find:

$$t^2+t-1=\left(t+\frac{1+\sqrt{5}}{2}\right)\left(t+\frac{1-\sqrt{5}}{2}\right)=\frac{1}{4}(2t+1+\sqrt{5})(2t+1-\sqrt{5})$$
 
karush said:
so then if
$$\displaystyle
t^2+t-1
=\left(t+\frac{1+\sqrt{5}}{2}\right)\left(t+\frac{1-\sqrt{5}}{2}\right)
=\frac{1}{4}(2t+1+\sqrt{5})(2t+1-\sqrt{5})$$
we can express the integral as
$$\displaystyle
I
=\frac{1}{4}\left[
\int\frac{t}{2t+1+\sqrt{5}}\ dt
+\int\frac{1}{2t+1-\sqrt{5}}\ dt \right]$$

No, check your partial fraction decomposition...you should get:

$$\frac{t+1}{t^2+t-1}=\frac{1}{\sqrt{5}}\left(\frac{\sqrt{5}-1}{2t+\sqrt{5}+1}+\frac{\sqrt{5}+1}{2t-\sqrt{5}+1}\right)$$
 
$\text{so if}$
$$\displaystyle \frac{t+1}{t^2+t-1}
=\frac{1}{\sqrt{5}}
\left(\frac{\sqrt{5}-1}{2t+\sqrt{5}+1}
+\frac{\sqrt{5}+1}{2t-\sqrt{5}+1}\right)$$
$\text{then}$
$$I=\frac{1}{\sqrt{5}}
\left[
\sqrt{5}-1 \int \frac{1}{2t+\sqrt{5}+1} \ dt
+\sqrt{5}+1\int\frac{1}{2t-\sqrt{5}-1} \ dt
\right]$$
$\text{ integrating}$
$$I=\frac{1}{\sqrt{5}}
\left[
(\sqrt{5}-1)(\ln({2t+\sqrt{5}+1}))
+(\sqrt{5}+1)\ln({2t-\sqrt{5}-1})
\right]+C$$
 
Last edited:
karush said:
$\text{so if}$
$$\displaystyle \frac{t+1}{t^2+t-1}
=\frac{1}{\sqrt{5}}
\left(\frac{\sqrt{5}-1}{2t+\sqrt{5}+1}
+\frac{\sqrt{5}+1}{2t-\sqrt{5}+1}\right)$$
$\text{then}$
$$I=\frac{1}{\sqrt{5}}
\left[
\sqrt{5}-1 \int \frac{1}{2t+\sqrt{5}+1} \ dt
+\sqrt{5}+1\int\frac{1}{2t-\sqrt{5}-1} \ dt
\right]$$
$\text{ integrating}$
$$I=\frac{1}{\sqrt{5}}
\left[
(\sqrt{5}-1)(\ln({2t+\sqrt{5}+1}))
+(\sqrt{5}+1)\ln({2t-\sqrt{5}-1})
\right]$$

That's not quite right...what is:

$$I=\int\frac{a}{2u+b}\,du$$?
 
MarkFL said:
That's not quite right...what is:

$$I=\int\frac{a}{2u+b}\,du$$?
$$
\displaystyle I=\int\frac{a}{2u+b}\,du
=\frac{a\ln\left({\left| 2u+b \right|}\right)}{2} +C$$

$\displaystyle
I=\frac{1}{\sqrt{5}}
\left[\frac
{(\sqrt{5}-1)(\ln({2t+\sqrt{5}+1}))}{2}

+\frac{(\sqrt{5}+1)\ln({2t-\sqrt{5}-1})}{2} \right]+C$

TA DA

$$\displaystyle
I=
+\left(\frac{5-\sqrt{5}}{10}\right)
\ln\left({2t+1+\sqrt{5}}\right)
+\left(\frac{5+\sqrt{5}}{10}\right)
\ln\left({2t+1-\sqrt{5}}\right) +C$$
 
Last edited:
karush said:
$\tiny{\text {Whitman 8.7.28 integral rational expression}} $
$$\displaystyle
\int\frac{t+1}{{t}^{2}+t-1}\ dt$$
$\text{book answer}$
$$\displaystyle\frac{5+\sqrt{5}}{10}
\ln\left({2t+1-\sqrt{5}}\right)
+\frac{5-\sqrt{5}}{10}
\ln\left({2t+1+\sqrt{5}}\right)+C$$

$\text{expansion}$
$$\displaystyle
\int\frac{t+1}{{t}^{2}+t-1}\ dt
=\int\frac{t}{{t}^{2}+t-1}\ dt
+\int\frac{1}{{t}^{2}+t-1}\ dt $$

Not sure how to approach this since it won't factor

An easier method perhaps?

$\displaystyle \begin{align*} \int{\frac{t + 1}{t^2 + t - 1}\,\mathrm{d}t} &= \frac{1}{2} \int{ \frac{2\,t + 2}{t^2 + t - 1}\,\mathrm{d}t } \\ &= \frac{1}{2} \int{ \frac{2\,t + 1}{t^2 + t - 1} \,\mathrm{d}t } + \frac{1}{2}\int{ \frac{1}{t^2 + t - 1} \,\mathrm{d}t } \\ &= \frac{1}{2} \int{ \frac{2\,t + 1}{t^2 + t - 1} \,\mathrm{d}t } + \frac{1}{2} \int{ \frac{1}{ t^2 + t + \left( \frac{1}{2} \right) ^2 - \left( \frac{1}{2} \right) ^2 - 1 } \,\mathrm{d}t } \\ &= \frac{1}{2} \int{ \frac{2\,t + 1}{t^2 + t - 1} \,\mathrm{d}t } + \frac{1}{2} \int{ \frac{1}{\left( t + \frac{1}{2} \right) ^2 - \frac{5}{4}}\,\mathrm{d}t } \end{align*}$

The first integral can be solved with a substitution $\displaystyle \begin{align*} u = t^2 + t - 1 \implies \mathrm{d}u = \left( 2\,t + 1 \right) \,\mathrm{d}t \end{align*}$ and the second can be solved with a substitution $\displaystyle \begin{align*} t + \frac{1}{2} = \frac{\sqrt{5}}{2}\,\cosh{(x)} \implies \mathrm{d}t = \frac{\sqrt{5}}{2}\,\sinh{(x)}\,\mathrm{d}x \end{align*}$.
 
https://drive.google.com/file/d/1iXnEH2ZmCMbPZeofOPsIKjGW2MdzlvEw/view?usp=sharing
 

Similar threads

Undergrad Different Substitution for Solving an Integral
  • · Replies 6 ·
Replies
6
Views
3K
High School I need to check if I am right solving this integral
  • · Replies 2 ·
Replies
2
Views
2K
Graduate How do I apply the method of steepest descents to this integral?
  • · Replies 1 ·
Replies
1
Views
3K
MHB 51 How far will the car travel in 10 sec
  • · Replies 2 ·
Replies
2
Views
2K
MHB What is the Integration by Parts method used for?
  • · Replies 2 ·
Replies
2
Views
1K
Undergrad Gaussian integral by differentiating under the integral sign
  • · Replies 19 ·
Replies
19
Views
5K
Undergrad Substitution in a definite integral
  • · Replies 6 ·
Replies
6
Views
3K
MHB Transforming Trigonometric Integrals with Substitution
  • · Replies 12 ·
Replies
12
Views
3K
MHB How Does Trig Substitution Simplify the Integral of 1/(25-t^2)^(3/2)?
  • · Replies 3 ·
Replies
3
Views
2K
Undergrad Integrating x^ne^xn: Analyzing & Solving
  • · Replies 5 ·
Replies
5
Views
4K