Integrate the function 5 x (x+1) (x^ 2 +9 )

Question

Integrate the function $\frac{5 x}{(x+1)\left(x^{2}+9\right)}$

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Answer

Given: $\frac{5 x}{(x+1)\left(x^{2}+9\right)} $
Let $I=\frac{5 x}{(x+1)\left(x^{2}+9\right)} $
Using partial fraction:
Let $\frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)} ...(i)$
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)} = \frac{A\left(x^{2}+9\right)+(B x+C)(x+1)}{(x+1)\left(x^{2}+9\right)} $
$\Rightarrow 5x = A(x^{2 }+ 9) + (Bx + C)(x + 1)$
$\Rightarrow 5x = Ax^2 + 9A + Bx^2 + Bx + Cx + C$
$\Rightarrow 5x = 9A + C + (B + C)x + (A + B)x^{2 }$
Equating the coefficients of $x, x^2$^ and constant value. We get:
$9A + C = 0 $
$\Rightarrow C = -9A$
$B+C = 5 $
$\Rightarrow B = 5 - C $
$\Rightarrow B = 5 - (-9A) $
$\Rightarrow B = 5 + 9A$
$A + B = 0 $
$\Rightarrow A = -B $
$\Rightarrow A = -(5 + 9A) $
$\Rightarrow 10A = -5 $
$\Rightarrow A = \frac{-1}{2} and C = \frac{9}{2} and B = \frac{1}{2} $
Put these values in equation $(i)$
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{(x+1)}+\frac{B x+C}{\left(x^{2}+9\right)} $
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=\frac{-\frac{1}{2}}{(x+1)}+\frac{\left(\frac{1}{2}\right) x+\frac{9}{2}}{\left(x^{2}+9\right)}$
$\Rightarrow \frac{5 x}{(x+1)\left(x^{2}+9\right)}=-\frac{1}{2} \cdot \frac{1}{(x+1)}+\frac{1}{2} \cdot\left(\frac{x+9}{\left(x^{2}+9\right)}\right) $
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x = -\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+\frac{1}{2} \cdot \int \frac{x}{\left(x^{2}+9\right)} d x+\frac{9}{2} \int \frac{1}{\left(x^{2}+9\right)} d x $
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x = -\frac{1}{2} \cdot \int \frac{1}{(x+1)} d x+I_{1}+\frac{9}{2} \int \frac{1}{\left(x^{2}+\left(3^{2}\right))\right.} d x $
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x = -\frac{1}{2} \cdot \log |\mathrm{x}+1|+\mathrm{I}_{1}+\frac{9}{2} \cdot\left(\frac{1}{3} \tan ^{-1} \frac{\mathrm{x}}{3}\right) ...(ii)$
First solve for $I_1:$
$I_{1}=\frac{1}{2} \cdot \int \frac{x}{\left(x^{2}+9\right)} d x $
$Put x^2 = t $
$\Rightarrow 2xdx = dt$
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2} \cdot \int \frac{1}{(\mathrm{t}+9)} \cdot \frac{\mathrm{dt}}{2} $
$\Rightarrow \mathrm{I}_{1}=\frac{1}{2} \cdot \int \frac{1}{(\mathrm{t}+9)} \cdot \frac{\mathrm{dt}}{2} $
$\Rightarrow \mathrm{I}_{1}=\frac{1}{4} \log \left|\mathrm{x}^{2}+9\right|$
Put the value in equ.$ (ii) $
$\Rightarrow \int \frac{5 x}{(x+1)\left(x^{2}+9\right)} d x = -\frac{1}{2} \cdot \log |\mathrm{x}+1|+\frac{1}{4} \log \left|\mathrm{x}^{2}+9\right|+\frac{3}{2} \cdot\left(\tan ^{-1} \frac{\mathrm{x}}{3}\right)+\mathrm{C}$

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