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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
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 ⇒ 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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