MHB Kishan's question via email about an indefinite integral

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The discussion focuses on solving the integral of the function (54t - 12) divided by the product of (t - 9) and (t^2 - 2). The recommended method is to use Partial Fractions to decompose the integrand into simpler fractions. The coefficients A, B, and C are determined through substitution, yielding A = 6, B = -3, and C = -3. The final integral solution is expressed in terms of logarithms, ensuring that absolute values are included for each logarithmic term. The solution is confirmed to be correct with the necessary modulus signs in place.
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What is the $\displaystyle \begin{align*} \int{ \frac{54\,t - 12}{\left( t- 9 \right) \left( t^2 - 2 \right) } \,\mathrm{d}t } \end{align*}$

We should use Partial Fractions to simplify the integrand. The denominator can be factorised further as $\displaystyle \begin{align*} \int{ \frac{54\,t - 12}{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } \end{align*}$, so that means the decomposition we should use is

$\displaystyle \begin{align*} \frac{A}{t - 9} + \frac{B}{t - \sqrt{2}} + \frac{C}{t + \sqrt{2}} &\equiv \frac{54\,t - 12}{\left( t - 9 \right) \left( t^2 - 2 \right) } \\ \frac{A\,\left( t - \sqrt{2} \right)\left( t + \sqrt{2} \right) + B\,\left( t - 9 \right) \left( t + \sqrt{2}\right) + C \,\left( t - 9 \right) \left( t - \sqrt{2} \right) }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } &\equiv \frac{54\,t - 12}{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \\ A\,\left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) + B \,\left( t - 9 \right) \left( t + \sqrt{2} \right) + C \,\left( t - 9 \right) \left( t - \sqrt{2} \right) &\equiv 54\,t - 12 \end{align*}$

Let $\displaystyle \begin{align*} t = 9 \end{align*}$ and we have $\displaystyle \begin{align*} 79\,A = 474 \implies A = 6 \end{align*}$.

Let $\displaystyle \begin{align*} t = \sqrt{2} \end{align*}$ and we have $\displaystyle \begin{align*} 2\,\sqrt{2} \,\left( \sqrt{2} - 9 \right)\,B = 54\,\sqrt{2} - 12 \implies \left( 4 - 18\,\sqrt{2} \right) \, B = 54\,\sqrt{2} - 12 \implies \left( 4 - 18\,\sqrt{2} \right) \, B = -3\,\left( 4 - 18\,\sqrt{2} \right) \implies B= -3 \end{align*}$.

Let $\displaystyle \begin{align*} t = -\sqrt{2} \end{align*}$ and we have $\displaystyle \begin{align*} -2\,\sqrt{2} \,\left( -\sqrt{2} - 9 \right) \, C = -54\,\sqrt{2} - 12 \implies \left( 4 + 18\,\sqrt{2} \right) \, C = -3\,\left( 4 + 18\,\sqrt{2} \right) \implies C = -3 \end{align*}$

So the integral becomes

$\displaystyle \begin{align*} \int{ \frac{54\,t - 12 }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } &= \int{ \left[ \frac{6}{t - 9} - \frac{3}{t - \sqrt{2}} - \frac{3}{t + \sqrt{2}} \right] \,\mathrm{d}t } \\ &= 6\ln{ \left| t - 9 \right| } - 3\ln{ \left| t - \sqrt{2} \right| } - 3\ln{ \left| t + \sqrt{2} \right| } + C \end{align*}$
 
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This is correct, however are
Prove It said:
What is the $\displaystyle \begin{align*} \int{ \frac{54\,t - 12}{\left( t- 9 \right) \left( t^2 - 2 \right) } \,\mathrm{d}t } \end{align*}$

We should use Partial Fractions to simplify the integrand. The denominator can be factorised further as $\displaystyle \begin{align*} \int{ \frac{54\,t - 12}{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } \end{align*}$, so that means the decomposition we should use is

$\displaystyle \begin{align*} \frac{A}{t - 9} + \frac{B}{t - \sqrt{2}} + \frac{C}{t + \sqrt{2}} &\equiv \frac{54\,t - 12}{\left( t - 9 \right) \left( t^2 - 2 \right) } \\ \frac{A\,\left( t - \sqrt{2} \right)\left( t + \sqrt{2} \right) + B\,\left( t - 9 \right) \left( t + \sqrt{2}\right) + C \,\left( t - 9 \right) \left( t - \sqrt{2} \right) }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } &\equiv \frac{54\,t - 12}{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \\ A\,\left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) + B \,\left( t - 9 \right) \left( t + \sqrt{2} \right) + C \,\left( t - 9 \right) \left( t - \sqrt{2} \right) &\equiv 54\,t - 12 \end{align*}$

Let $\displaystyle \begin{align*} t = 9 \end{align*}$ and we have $\displaystyle \begin{align*} 79\,A = 474 \implies A = 6 \end{align*}$.

Let $\displaystyle \begin{align*} t = \sqrt{2} \end{align*}$ and we have $\displaystyle \begin{align*} 2\,\sqrt{2} \,\left( \sqrt{2} - 9 \right)\,B = 54\,\sqrt{2} - 12 \implies \left( 4 - 18\,\sqrt{2} \right) \, B = 54\,\sqrt{2} - 12 \implies \left( 4 - 18\,\sqrt{2} \right) \, B = -3\,\left( 4 - 18\,\sqrt{2} \right) \implies B= -3 \end{align*}$.

Let $\displaystyle \begin{align*} t = -\sqrt{2} \end{align*}$ and we have $\displaystyle \begin{align*} -2\,\sqrt{2} \,\left( -\sqrt{2} - 9 \right) \, C = -54\,\sqrt{2} - 12 \implies \left( 4 + 18\,\sqrt{2} \right) \, C = -3\,\left( 4 + 18\,\sqrt{2} \right) \implies C = -3 \end{align*}$

So the integral becomes

$\displaystyle \begin{align*} \int{ \frac{54\,t - 12 }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } &= \int{ \left[ \frac{6}{t - 9} - \frac{3}{t - \sqrt{2}} - \frac{3}{t + \sqrt{2}} \right] \,\mathrm{d}t } \\ &= 6\ln{ \left| t - 9 \right| } - 3\ln{ \left| t - \sqrt{2} \right| } - 3\ln{ \left| t + \sqrt{2} \right| } + C \end{align*}$

This is correct, however are you not missing the modulus sign on the last part of your solution? Is it not supposed to be;

$\displaystyle \begin{align*} \int{ \frac{54\,t - 12 }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } &= \int{ \left[ \frac{6}{t - 9} - \frac{3}{t - \sqrt{2}} - \frac{3}{t + \sqrt{2}} \right] \,\mathrm{d}t } \\ &= 6\ln{ \left| t - 9 \right| } - 3\ln{ \left|| t - \sqrt{2} \right|| } - 3\ln{ \left|| t + \sqrt{2} \right|| } + C \end{align*}$
 
chwala said:
This is correct, however are you not missing the modulus sign on the last part of your solution? Is it not supposed to be;

$\displaystyle \begin{align*} \int{ \frac{54\,t - 12 }{ \left( t - 9 \right) \left( t - \sqrt{2} \right) \left( t + \sqrt{2} \right) } \,\mathrm{d}t } &= \int{ \left[ \frac{6}{t - 9} - \frac{3}{t - \sqrt{2}} - \frac{3}{t + \sqrt{2}} \right] \,\mathrm{d}t } \\ &= 6\ln{ \left| t - 9 \right| } - 3\ln{ \left|| t - \sqrt{2} \right|| } - 3\ln{ \left|| t + \sqrt{2} \right|| } + C \end{align*}$
The given solution does have absolute value in each logarithm.

##\displaystyle 6\ln{ \left| \,t - 9\, \right| } - 3\ln{ \left| \, t - \sqrt{2} \, \right| } - 3\ln{ \left| \, t + \sqrt{2} \,\right| } + C ##
 
SammyS said:
The given solution does have absolute value in each logarithm.

##\displaystyle 6\ln{ \left| \,t - 9\, \right| } - 3\ln{ \left| \, t - \sqrt{2} \, \right| } - 3\ln{ \left| \, t + \sqrt{2} \,\right| } + C ##
Noted @SammyS
 
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