MHB -w8.7.28 integral rational expression

[karush]

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

 

[MarkFL]

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:

$$t^2+t-1=\left(t+\frac{1+\sqrt{5}}{2}\right)\left(t+\frac{1-\sqrt{5}}{2}\right)=\frac{1}{4}(2t+1+\sqrt{5})(2t+1-\sqrt{5})$$

 

[MarkFL]

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:

$$\frac{t+1}{t^2+t-1}=\frac{1}{\sqrt{5}}\left(\frac{\sqrt{5}-1}{2t+\sqrt{5}+1}+\frac{\sqrt{5}+1}{2t-\sqrt{5}+1}\right)$$

 

[karush]

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

 

[MarkFL]

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

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

$$I=\int\frac{a}{2u+b}\,du$$?

 

[karush]

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

$$\displaystyle
I=
+\left(\frac{5-\sqrt{5}}{10}\right)
\ln\left({2t+1+\sqrt{5}}\right)
+\left(\frac{5+\sqrt{5}}{10}\right)
\ln\left({2t+1-\sqrt{5}}\right) +C$$

 

[Prove It]

$\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 $\displaystyle \begin{align*} t + \frac{1}{2} = \frac{\sqrt{5}}{2}\,\cosh{(x)} \implies \mathrm{d}t = \frac{\sqrt{5}}{2}\,\sinh{(x)}\,\mathrm{d}x \end{align*}$.

 

[karush]

https://drive.google.com/file/d/1iXnEH2ZmCMbPZeofOPsIKjGW2MdzlvEw/view?usp=sharing