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Kishan's question via email about an indefinite integral

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
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

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