Question

If $\int \frac{1}{x^4+8x^2+9}dx = \frac{1}{k}\left[ \frac{1}{\sqrt{14}}\tan^{-1}(f(x)) - \frac{1}{\sqrt{2}}\tan^{-1}(g(x))\right] + c,$

then $\sqrt{\frac{k}{2}+ f(\sqrt{3}) + g(1)} =$

Answer 1

$\sqrt{2}+1$

Answer 2

The integral we need to solve is $I = \int \frac{1}{x^4+8x^2+9}dx$. This is a standard type of integral involving $x^4$. We can divide the numerator and denominator by $x^2$. $I = \int \frac{1/x^2}{x^2+8+9/x^2}dx$.

We can rewrite the denominator using the identity $x^2 + \frac{a^2}{x^2} = (x \pm \frac{a}{x})^2 \mp 2a$. Here, $a=3$. So, $x^2 + \frac{9}{x^2} = (x - \frac{3}{x})^2 + 6$ or $x^2 + \frac{9}{x^2} = (x + \frac{3}{x})^2 - 6$. The denominator is $x^2+9/x^2+8$. Using the first identity: $x^2+9/x^2+8 = (x-3/x)^2 + 6 + 8 = (x-3/x)^2 + 14$. Using the second identity: $x^2+9/x^2+8 = (x+3/x)^2 - 6 + 8 = (x+3/x)^2 + 2$.

The numerator is $1/x^2$. The derivative of $(x-3/x)$ is $1+3/x^2$, and the derivative of $(x+3/x)$ is $1-3/x^2$. We can express $1/x^2$ as a linear combination of these derivatives: $\frac{1}{x^2} = A\left(1+\frac{3}{x^2}\right) + B\left(1-\frac{3}{x^2}\right)$ $\frac{1}{x^2} = (A+B) + \frac{3(A-B)}{x^2}$. Comparing coefficients of $1$ and $1/x^2$: $A+B = 0$ $3(A-B) = 1 \implies A-B = 1/3$. Adding the two equations: $2A = 1/3 \implies A = 1/6$. Subtracting the second from the first: $2B = -1/3 \implies B = -1/6$. So, $\frac{1}{x^2} = \frac{1}{6}\left(1+\frac{3}{x^2}\right) - \frac{1}{6}\left(1-\frac{3}{x^2}\right)$.

Substitute this back into the integral: $I = \int \frac{\frac{1}{6}\left(1+\frac{3}{x^2}\right) - \frac{1}{6}\left(1-\frac{3}{x^2}\right)}{x^2+9/x^2+8}dx$ $I = \frac{1}{6} \int \frac{1+3/x^2}{(x-3/x)^2+14}dx - \frac{1}{6} \int \frac{1-3/x^2}{(x+3/x)^2+2}dx$.

For the first integral, let $u = x - 3/x$. Then $du = (1+3/x^2)dx$. The integral becomes $\int \frac{du}{u^2+(\sqrt{14})^2}$. $\int \frac{du}{u^2+(\sqrt{14})^2} = \frac{1}{\sqrt{14}}\tan^{-1}\left(\frac{u}{\sqrt{14}}\right) = \frac{1}{\sqrt{14}}\tan^{-1}\left(\frac{x-3/x}{\sqrt{14}}\right) = \frac{1}{\sqrt{14}}\tan^{-1}\left(\frac{x^2-3}{x\sqrt{14}}\right)$.

For the second integral, let $v = x + 3/x$. Then $dv = (1-3/x^2)dx$. The integral becomes $\int \frac{dv}{v^2+(\sqrt{2})^2}$. $\int \frac{dv}{v^2+(\sqrt{2})^2} = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{v}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x+3/x}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^2+3}{x\sqrt{2}}\right)$.

Combining the results, we get: $I = \frac{1}{6} \left[ \frac{1}{\sqrt{14}}\tan^{-1}\left(\frac{x^2-3}{x\sqrt{14}}\right) - \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x^2+3}{x\sqrt{2}}\right) \right] + c$.

The given form is $\frac{1}{k}\left[ \frac{1}{\sqrt{14}}\tan^{-1}(f(x)) - \frac{1}{\sqrt{2}}\tan^{-1}(g(x))\right] + c$. Comparing the two forms, we identify: $k = 6$. $f(x) = \frac{x^2-3}{x\sqrt{14}}$. $g(x) = \frac{x^2+3}{x\sqrt{2}}$.

We need to evaluate the expression $\sqrt{\frac{k}{2}+ f(\sqrt{3}) + g(1)}$. First, calculate the required values: $k = 6$. $f(\sqrt{3}) = \frac{(\sqrt{3})^2-3}{\sqrt{3}\sqrt{14}} = \frac{3-3}{\sqrt{42}} = \frac{0}{\sqrt{42}} = 0$. $g(1) = \frac{(1)^2+3}{1\sqrt{2}} = \frac{1+3}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$.

Now substitute these values into the expression: $\sqrt{\frac{k}{2}+ f(\sqrt{3}) + g(1)} = \sqrt{\frac{6}{2}+ 0 + 2\sqrt{2}} = \sqrt{3 + 2\sqrt{2}}$.

To simplify $\sqrt{3 + 2\sqrt{2}}$, we look for two numbers whose sum is 3 and product is 2. These numbers are 2 and 1. So, $3 + 2\sqrt{2} = (\sqrt{2})^2 + (1)^2 + 2 \cdot \sqrt{2} \cdot 1 = (\sqrt{2}+1)^2$. $\sqrt{3 + 2\sqrt{2}} = \sqrt{(\sqrt{2}+1)^2} = |\sqrt{2}+1|$. Since $\sqrt{2}+1$ is positive, $|\sqrt{2}+1| = \sqrt{2}+1$.

The value of the expression is $\sqrt{2}+1$.