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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Find the integral: $\int \frac{d x}{3 x^{2}+13 x-10}$
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Answer

We write the denominator of the integrand,
$3x^2 + 13x - 10 = 3\left(x^{2}+\frac{13 x}{3}-\frac{10}{3}\right)$
$= 3\left[\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}\right] ($completing the square$)$
Thus, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}\right)^{2}-\left(\frac{17}{6}\right)^{2}}$ 
Put $x+\frac{13}{6}=t$. Then $dx = dt$
Therefore, $\int \frac{d x}{3 x^{2}+13 x-10}=\frac{1}{3} \int \frac{d t}{t^{2}-\left(\frac{17}{6}\right)^{2}}$ 
$= \frac{1}{3 \times 2 \times \frac{17}{6}} \log \left|\frac{t-\frac{17}{6}}{t+\frac{17}{6}}\right|+\mathrm{C}_{1}$ 
$= \frac{1}{17} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}\right|+C_{1}$ 
$= \frac{1}{17} \log \left|\frac{6 x-4}{6 x+30}\right|+C_{1}$ 
$= \frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C_{1}+\frac{1}{17} \log \frac{1}{3}$ 
$= \frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C$ where, $\mathrm{C}=\mathrm{C}_{1}+\frac{1}{17} \log \frac{1}{3}$

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