Find the following integral: int frac{d x}{3 | vedclass

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Question

(A) The given integral is of the form $\int \frac{d x}{a x^{2}+b x+c}$. We write the denominator of the integrand as:$3 x^{2}+13 x-10 = 3\left(x^{2}+\

Vedclass · Accepted Answer

$\frac{1}{17} \log \left|\frac{3 x-2}{x+5}\right|+C$

Vedclass · Answer

(A) The given integral is of the form $\int \frac{d x}{a x^{2}+b x+c}$. We write the denominator of the integrand as:$3 x^{2}+13 x-10 = 3\left(x^{2}+\frac{13}{3}x-\frac{10}{3}ight)$$= 3\left[\left(x+\frac{13}{6}ight)^{2}-\left(\frac{13}{6}ight)^{2}-\frac{10}{3}ight]$$= 3\left[\left(x+\frac{13}{6}ight)^{2}-\frac{169}{36}-\frac{120}{36}ight] = 3\left[\left(x+\frac{13}{6}ight)^{2}-\left(\frac{17}{6}ight)^{2}ight]$Thus,$\int \frac{d x}{3 x^{2}+13 x-10} = \frac{1}{3} \int \frac{d x}{\left(x+\frac{13}{6}ight)^{2}-\left(\frac{17}{6}ight)^{2}}$Using the formula $\int \frac{dx}{x^2-a^2} = \frac{1}{2a} \log \left|\frac{x-a}{x+a}ight| + C$,we get:$= \frac{1}{3} 	imes \frac{1}{2 	imes \frac{17}{6}} \log \left|\frac{x+\frac{13}{6}-\frac{17}{6}}{x+\frac{13}{6}+\frac{17}{6}}ight| + C$$= \frac{1}{17} \log \left|\frac{x-\frac{4}{6}}{x+\frac{30}{6}}ight| + C = \frac{1}{17} \log \left|\frac{x-\frac{2}{3}}{x+5}ight| + C$$= \frac{1}{17} \log \left|\frac{3x-2}{3(x+5)}ight| + C = \frac{1}{17} \log \left|\frac{3x-2}{x+5}ight| + C'$

Find the following integral: $\int \frac{d x}{3 x^{2}+13 x-10}$

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