If a 0, b 0 and int frac{1}{ax^2+b} dx = | vedclass

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(B) The standard integral is $\int \frac{1}{Ax^2+B} dx = \frac{1}{\sqrt{AB}} 	an^{-1} \left(x \sqrt{\frac{A}{B}}ight) + c$.Given $\int \frac{1}{ax^

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$\frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{\sqrt{3}x}{\sqrt{2}}ight) + c$

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(B) The standard integral is $\int \frac{1}{Ax^2+B} dx = \frac{1}{\sqrt{AB}} 	an^{-1} \left(x \sqrt{\frac{A}{B}}ight) + c$.Given $\int \frac{1}{ax^2+b} dx = \frac{1}{\sqrt{ab}} 	an^{-1} \left(x \sqrt{\frac{a}{b}}ight) + c = \frac{1}{\sqrt{6}} 	an^{-1} \left(x \sqrt{\frac{2}{3}}ight) + c$.Comparing, we get $\sqrt{ab} = \sqrt{6} \implies ab = 6$ and $\sqrt{\frac{a}{b}} = \sqrt{\frac{2}{3}} \implies \frac{a}{b} = \frac{2}{3}$.Solving $ab=6$ and $a = \frac{2b}{3}$, we get $(\frac{2b}{3})b = 6 \implies b^2 = 9 \implies b = 3$ and $a = 2$.Now, $\int \frac{1}{bx^2+a} dx = \int \frac{1}{3x^2+2} dx = \frac{1}{\sqrt{3 \cdot 2}} 	an^{-1} \left(x \sqrt{\frac{3}{2}}ight) + c = \frac{1}{\sqrt{6}} 	an^{-1} \left(\frac{\sqrt{3}x}{\sqrt{2}}ight) + c$.

If $a > 0, b > 0$ and $\int \frac{1}{ax^2+b} dx = \frac{1}{\sqrt{6}} \tan^{-1} \left(\frac{\sqrt{2}x}{\sqrt{3}}\right) + c$, then $\int \frac{1}{bx^2+a} dx = \dots$

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