Evaluate $\displaystyle \int \dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} \; dx$
Let $I = \displaystyle \int \dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} \; dx$ $\;\;\; \cdots \; (1)$
Let $\dfrac{5x}{\left(x + 1\right) \left(x^2 + 9\right)} = \dfrac{A}{x + 1} + \dfrac{Bx + C}{x^2 + 9}$ $\;\;\; \cdots \; (2)$
$\begin{aligned}
\text{i.e. } \; 5x & = A \left(x^2 + 9\right) + \left(Bx + C\right) \left(x + 1\right) \\\\
& = A x^2 + 9 A + B x^2 + B x + C x + C \\\\
& = x^2 \left(A + B\right) + x \left(B + C\right) + \left(9 A + C\right)
\end{aligned}$
Comparing the coefficient of $x^2$ term gives
$A + B = 0$ $\implies$ $B = - A$ $\;\;\; \cdots \; (3a)$
Comparing the coefficient of $x$ term gives
$B + C = 5$ $\implies$ $- A + C = 5$ $\;\;\; \cdots \; (3b)$ [from equation $(3a)$]
Comparing the constant term gives
$9 A + C = 0$ $\;\;\; \cdots \; (3c)$
Subtracting equations $(3b)$ and $(3c)$ gives
$10 A = - 5$ $\implies$ $A = \dfrac{-1}{2}$
$\therefore$ $\;$ We have from equation $(3a)$, $B = -A = \dfrac{1}{2}$
From equation $(3b)$, $C = 5 + A = 5 - \dfrac{1}{2} = \dfrac{9}{2}$
Substituting the values of A, B and C in equation $(2)$ gives
$\dfrac{5 x}{\left(x + 1\right) \left(x^2 + 9\right)} = \dfrac{-1}{2 \left(x + 1\right)} + \dfrac{\dfrac{1}{2}x + \dfrac{9}{2}}{x^2 + 9}$ $\;\;\; \cdots \; (4)$
In view of equation $(4)$, equation $(1)$ becomes
$I = \dfrac{-1}{2} \displaystyle \int \dfrac{dx}{x + 1} + \dfrac{1}{2} \displaystyle \int \dfrac{x}{x^2 + 9} \; dx + \dfrac{9}{2} \displaystyle \int \dfrac{dx}{x^2 + 9}$ $\;\;\; \cdots \; (5)$
Now, $\displaystyle \int \dfrac{dx}{x + 1} = \log \left|x + 1\right| + c_1$ $\;\;\; \cdots \; (6a)$
Consider $\displaystyle \int \dfrac{x}{x^2 + 9} \; dx$
Let $x^2 + 9 = t$ $\;\;\; \cdots \; (7a)$
Differentiating equation $(7a)$ w.r.t x gives
$2 x \; dx = dt$ $\implies$ $x \; dx = \dfrac{dt}{2}$ $\;\;\; \cdots \; (7b)$
$\therefore$ $\;$ In view of equations $(7a)$ and $(7b)$ we have
$\begin{aligned}
\int \dfrac{x \; dx}{x^2 + 9} & = \dfrac{1}{2} \int \dfrac{dt}{t} \\\\
& = \dfrac{1}{2} \log \left|t\right| + c_2 \\\\
& = \dfrac{1}{2} \log \left|x^2 + 9\right| + c_2 \;\;\; \left[\text{from equation (7a)}\right] \;\;\; \cdots (6b)
\end{aligned}$
$\begin{aligned}
\text{Consider } \int \dfrac{dx}{x^2 + 9} & = \int \dfrac{dx}{\left(x\right)^2 + \left(3\right)^2} \\\\
& = \dfrac{1}{3} \tan^{-1} \left(\dfrac{x}{3}\right) + c_3 \;\;\; \cdots \; (6c) \\\\
& \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c\right]
\end{aligned}$
$\therefore$ $\;$ In view of equations $(6a)$, $(6b)$ and $(6c)$, equation $(5)$ becomes
$I = \dfrac{-1}{2} \log \left|x + 1\right| + \dfrac{1}{4} \log \left|x^2 + 9\right| + \dfrac{3}{2} \tan^{-1} \left(\dfrac{x}{3}\right) + c$
where $c = \dfrac{- c_1}{2} + \dfrac{c_2}{2} + \dfrac{9 c_3}{2}$