91Ó°ÊÓ

The given substitution is \(x = (u^2 - 1)^2\). First, find the derivative of \(x\) with respect to \(u\): $$\frac{dx}{du} = \frac{d}{du}((u^2-1)^2)$$ Now, differentiate the function using the chain rule: $$\frac{dx}{du} = 2(u^2-1) \cdot 2u$$ $$\frac{dx}{du} = 4u (u^2-1)$$ Now, to make the substitution and change the variables from \(x\) to \(u\), we need to express \(\frac{dx}{du}\) as \(\frac{du}{dx}\): $$\frac{du}{dx} = \frac{1}{4u(u^2-1)}$$ Substitute \(x\) and \(\frac{du}{dx}\) in the given integral: $$\int \frac{1}{\sqrt{1+\sqrt{x}}} dx = \int \frac{1}{\sqrt{1+\sqrt{(u^2-1)^2}}}\frac{du}{dx}dx$$ This simplifies to: $$\int \frac{1}{\sqrt{1+\sqrt{(u^2-1)^2}}} \frac{1}{4u(u^2-1)} du$$

Now, we need to simplify the integral. Let's focus on the denominator of the integrand. Note that: $$\sqrt{1+\sqrt{x}} = \sqrt{1+\sqrt{(u^2-1)^2}}$$ $$\sqrt{1+\sqrt{x}} = \sqrt{1+|u^2-1|}$$ We can assume that \(u^2>1\) (as the variable \(u\) is either larger or smaller than \(1\)), so the denominator becomes: $$\sqrt{1+u^2-1} = u$$ Now the integral simplifies to: $$\int \frac{1}{4u(u^2-1)} du$$

This integral is a rational function, which can be integrated using partial fraction decomposition. We rewrite the integrand as: $$\frac{1}{4u(u^2-1)} = \frac{A}{u} + \frac{Bu+C}{u^2-1}$$ Multiply both sides by the common denominator \(4u(u^2-1)\) to eliminate the denominators: $$1 = A(u^2-1) + (Bu+C)u$$ Setting \(u=0\), we find that \(A=-\frac{1}{2}\). Then, we can write the remaining equation as: $$1 = -\frac{1}{2}(u^2-1) + (Bu+C)u$$ $$1 = -(u^2/2)+\frac{1}{2}+(Bu+C)u = (Bu+C)u-(u^2/2)+\frac{1}{2}$$ Comparing coefficients, we get the following system of equations: $$B+\frac{1}{2}=0 \quad \Rightarrow \quad B=-\frac{1}{2}$$ $$C=\frac{1}{2}$$ Thus, our partial fraction decomposition is: $$\frac{1}{4u(u^2-1)} = -\frac{1}{2u} - \frac{u-1}{2(u^2-1)}$$ Now, we can integrate each term separately: $$\int \frac{1}{4u(u^2-1)} du = -\frac{1}{2}\int\frac{1}{u}du - \frac{1}{2}\int\frac{u-1}{(u^2-1)}du$$ Integration leads to: $$-\frac{1}{2}\ln|u| - \frac{1}{2}\ln|u^2-1|+C$$ Finally, we substitute the original variable \(x\) back into the result to get the final answer: $$-\frac{1}{2}\ln|\sqrt{x}+x-1| - \frac{1}{2}\ln|(\sqrt{x}+x-1)(\sqrt{x}-1)|+C$$ The final result is: $$\int \frac{dx}{\sqrt{1+\sqrt{x}}} = -\frac{1}{2}\ln|\sqrt{x}+x-1| - \frac{1}{2}\ln|(\sqrt{x}+x-1)(\sqrt{x}-1)|+C$$