If x neq frac{-3}{sqrt{2}},then int frac{x^2} | vedclass

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(A) Given,$\int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x$. Note that $2 x^2+6 \sqrt{2} x+9 = (\sqrt{2} x+3)^2$. We can write the numerator as: $x^2 = \fra

Vedclass · Accepted Answer

$\frac{1}{2 \sqrt{2}}\left[(\sqrt{2} x+3)-6 \log |\sqrt{2} x+3|-\frac{9}{\sqrt{2} x+3}\right]+c$

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(A) Given,$\int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x$. Note that $2 x^2+6 \sqrt{2} x+9 = (\sqrt{2} x+3)^2$. We can write the numerator as: $x^2 = \frac{1}{2} (2 x^2 + 6 \sqrt{2} x + 9) - 3 \sqrt{2} x - \frac{9}{2}$. So,the integral becomes: $\int \left( \frac{1}{2} - \frac{3 \sqrt{2} x + \frac{9}{2}}{(\sqrt{2} x + 3)^2} \right) d x = \frac{x}{2} - \int \frac{3 \sqrt{2} x + \frac{9}{2}}{(\sqrt{2} x + 3)^2} d x$. Let $u = \sqrt{2} x + 3$,then $du = \sqrt{2} dx$,so $dx = \frac{du}{\sqrt{2}}$ and $x = \frac{u-3}{\sqrt{2}}$. The integral part is $\int \frac{3 \sqrt{2} (\frac{u-3}{\sqrt{2}}) + \frac{9}{2}}{u^2} \frac{du}{\sqrt{2}} = \frac{1}{\sqrt{2}} \int \frac{3u - 9 + \frac{9}{2}}{u^2} du = \frac{1}{\sqrt{2}} \int \frac{3u - \frac{9}{2}}{u^2} du = \frac{1}{\sqrt{2}} \int \left( \frac{3}{u} - \frac{9}{2u^2} \right) du$. $= \frac{1}{\sqrt{2}} [3 \log |u| + \frac{9}{2u}] + c = \frac{3}{\sqrt{2}} \log |\sqrt{2} x + 3| + \frac{9}{2 \sqrt{2} (\sqrt{2} x + 3)} + c$. Substituting back into the expression: $\frac{x}{2} - \left( \frac{3}{\sqrt{2}} \log |\sqrt{2} x + 3| + \frac{9}{2 \sqrt{2} (\sqrt{2} x + 3)} \right) + c$. To match the form in the options,we factor out $\frac{1}{2 \sqrt{2}}$: $= \frac{1}{2 \sqrt{2}} [\sqrt{2} x - 6 \log |\sqrt{2} x + 3| - \frac{9}{\sqrt{2} x + 3}] + c$. Adding and subtracting $3$ inside the bracket: $= \frac{1}{2 \sqrt{2}} [(\sqrt{2} x + 3) - 3 - 6 \log |\sqrt{2} x + 3| - \frac{9}{\sqrt{2} x + 3}] + c$. Since the constant term can be absorbed into $c$,this matches option $(a)$.

If $x \neq \frac{-3}{\sqrt{2}}$,then $\int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x=$

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