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

Question

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

Aicharya · Accepted Answer

Given, $ \begin{aligned} & \int \frac{x^2}{2 x^2+6 \sqrt{2} x+9} d x \\ & \int \frac{\frac{1}{2}\left(2 x^2+6 \sqrt{2} x+9\right)-\left(3 \sqrt{2} x+\frac{9}{2}\right)}{2 x^2+6 \sqrt{2} x+9} d x \\ & {\left[\therefore x^2=\left(2 x^2+6 \sqrt{2} x+9\right) \frac{1}{2}-\left(3 \sqrt{2} x+\frac{9}{2}\right)\right]} \\ & \int\left(\frac{1}{2}-\frac{\left(3 \sqrt{2} x+\frac{9}{2}\right)}{2 x^2+6 \sqrt{2} x+9}\right) d x \\ & \int \frac{1}{2} d x-\int \frac{3 \sqrt{2} x+\frac{9}{2}}{2 x^2+6 \sqrt{2} x+9} d x \\ & \end{aligned} $

$3 \sqrt{2} x+\frac{9}{2}=A \frac{d}{d x}\left(2 x^2+6 \sqrt{2} x+9\right)+B$

On comparison, we get $ \begin{aligned} & 4 A=3 \sqrt{2} \\ & 6 \sqrt{2} A+B=\frac{9}{2} \\ & A=\frac{3 \sqrt{2}}{4} \\ & 6 \sqrt{2} \times \frac{3 \sqrt{2}}{4}+B=\frac{9}{2} \\ & B=-\frac{9}{2} \end{aligned} $ Substituting values of $A, B$ Eq. (ii)

Substituting Eq. (iii) in Eq. (i) $ \begin{gathered} \frac{x}{2}-\int \frac{\left(\frac{3 \sqrt{2}}{4}(4 x+6 \sqrt{2})-\frac{9}{2}\right) d x}{2 x^2+6 \sqrt{2} x+9} \\ \frac{x}{2}-\frac{3 \sqrt{2}}{4} \int \frac{4 x+6 \sqrt{2} d x}{2 x^2+6 \sqrt{2} x+9}+\frac{9}{2} \int \frac{1}{2 x^2+6 \sqrt{2} x+9} d x \\ \frac{x}{2}-\frac{3 \sqrt{2}}{4} \log \left|2 x^2+6 \sqrt{2} x+9\right|+\frac{9}{2} \int \frac{1}{(\sqrt{2} x+3)^2} d x \\ \\ {\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|\right]} \end{gathered} $ $ \begin{aligned} & \frac{x}{2}-\frac{3 \sqrt{2}}{4} \log \left|(\sqrt{2} x+3)^2\right|+\frac{9}{2} \frac{(\sqrt{2} x+3)^{-2+1}}{-2+1} \times \frac{1}{\sqrt{2}}+c^{\prime} \\ & \frac{x}{2}-\frac{3 \sqrt{2}}{4} \times 2 \log |\sqrt{2} x+3|+\frac{9}{2} \frac{(\sqrt{2} x+3)^{-1}}{-1} \times \frac{1}{\sqrt{2}}+c^{\prime} \\ & \frac{x}{2}-\frac{3 \sqrt{2}}{2} \log |\sqrt{2} x+3|-\frac{9}{2 \sqrt{2}} \times \frac{1}{\sqrt{2} x+3}+c^{\prime} \\ & \frac{1}{2 \sqrt{2}}\left[\sqrt{2} x-6 \log |\sqrt{2} x+3|-\frac{9}{\sqrt{2} x+3}\right]+c^{\prime} \\ & =\frac{1}{2 \sqrt{2}}\left[(\sqrt{2} x+3)-6 \log |\sqrt{2} x+3|-\frac{9}{\sqrt{2} x+3}\right]+c \end{aligned} $ Hence, option (a) is correct

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

Solution