# How do I find the integral int10/((x-1)(x^2+9))dx ?

Jul 30, 2014

$= \frac{5}{4} \ln \left(x - 1\right) - \frac{5}{8} \ln \left({x}^{2} + 9\right) + \frac{5}{12} {\tan}^{-} 1 \left(\frac{x}{3}\right) + c$, where $c$ is a constant

Explanation :

This type of question usually solve by using Partial Fractions,

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)}$, it can be written as

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 9}$

multiplying both sides by $\left(x - 1\right) \left({x}^{2} + 9\right)$, we get

$10 = A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x - 1\right)$

$10 = \left(A + B\right) {x}^{2} + \left(C - B\right) x + \left(9 A - C\right)$

comparing coefficients of ${x}^{2}$, $x$ and constant both side,

$A + B = 0$ $\implies$$A = - B$ .................$\left(i\right)$

$C - B = 0$$\implies$$C = - B$ .................$\left(i i\right)$

$9 A - C = 10$$\implies$ .................$\left(i i i\right)$

substituting $A$ & $C$from $\left(i\right)$ and $\left(i i\right)$ in $\left(i i i\right)$

$9 \left(- B\right) - \left(- B\right) = 10$

$B = - \frac{10}{8} = - \frac{5}{4}$

Therefore, $A = C = \frac{5}{4}$

Hence, we get

$\frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} = \frac{5}{4 \left(x - 1\right)} + \frac{\left(- \frac{5}{4}\right) x + \left(\frac{5}{4}\right)}{{x}^{2} + 9}$

integrating both sides with respect to $x$,

$\int \frac{10}{\left(x - 1\right) \left({x}^{2} + 9\right)} \mathrm{dx} = \int \frac{5}{4 \left(x - 1\right)} \mathrm{dx} + \int \frac{\left(- \frac{5}{4}\right) x + \left(\frac{5}{4}\right)}{{x}^{2} + 9} \mathrm{dx}$

$= \frac{5}{4} \int \frac{1}{\left(x - 1\right)} \mathrm{dx} - \frac{5}{4} \int \frac{x}{{x}^{2} + 9} \mathrm{dx} + \frac{5}{4} \int \frac{1}{{x}^{2} + 9} \mathrm{dx}$

Second integral is solved by substitution, as the numerator is the differentiation of denominator

$= \frac{5}{4} \ln \left(x - 1\right) - \frac{5}{4} \cdot \frac{1}{2} \ln \left({x}^{2} + 9\right) + \frac{5}{4} \cdot \frac{1}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right)$

$= \frac{5}{4} \ln \left(x - 1\right) - \frac{5}{8} \ln \left({x}^{2} + 9\right) + \frac{5}{12} {\tan}^{-} 1 \left(\frac{x}{3}\right) + c$, where $c$ is a constant