91Ó°ÊÓ

First, we complete the square for the expression inside the square root. $$x^2+10x = (x^2+10x+25) - 25 = (x+5)^2 - 25$$ Thus, the integral becomes: $$\int \sqrt{(x+5)^2 - 25} dx$$

To simplify the integral, we can use a substitution of the form \(x+5 = A\cdot \cosh(u)\), where \(A\) is a constant to be determined. We choose this substitution because the square root of a square minus a constant is usually simplified using the hyperbolic cosine function. Using the identities \(\cosh^2(u) - \sinh^2(u) = 1\) and \(\cosh^2(u) - 1 = \sinh^2(u)\), we can see that if we want \((x+5)^2 - 25\) to be a perfect square, we must have that \(A^2 - 25 = 25\). Solving this equation for \(A\), we find that \(A = 5\). Therefore, our substitution will be: $$x+5 = 5\cosh(u)$$

Differentiating our substitution, \(x+5 = 5\cosh(u)\), with respect to \(u\), we get: $$\frac{dx}{du} = 5\sinh(u)$$ Thus, \(dx = 5\sinh(u) du\).

Now, we substitute our expressions for \(x+5\) and \(dx\) into the integral and simplify: \begin{align*} \int \sqrt{(x+5)^2 - 25} dx &= \int \sqrt{25(\cosh^2(u)-1)} \cdot 5\sinh(u) du \\ &= 25 \int \sinh(u)\cdot \sqrt{\sinh^2(u)} du \\ &= 25 \int \sinh(u) \cdot \sinh(u) du \\ &= 25 \int \sinh^2(u) du \end{align*}

Now, we can use a table of integrals to find the antiderivative of \(\sinh^2(u)\). Using the integral for \(\sinh^2(u)\), we have: $$\int \sinh^2(u) du = \frac{1}{2}(u-\sinh(u)\cosh(u)) + C$$

Now, we must undo our substitution and rewrite the integral in terms of \(x\). Recall that \(x+5 = 5\cosh(u)\). We can rewrite this in terms of \(\cosh(u)\) as follows: $$\cosh(u) = \frac{x+5}{5}$$ To find \(\sinh(u)\), we can use the identity \(\sinh^2(u) = \cosh^2(u) - 1\): $$\sinh(u) = \sqrt{\cosh^2(u) - 1} = \frac{\sqrt{(x+5)^2 - 25}}{5}$$ We also need to find \(u\) in terms of \(x\). To do this, we can take the inverse hyperbolic cosine function on both sides of our equation \(\cosh(u) = \frac{x+5}{5}\): $$u = \cosh^{-1}\left(\frac{x+5}{5}\right)$$ Substituting these expressions back into our antiderivative, we get: $$\int \sqrt{x^2+10x} dx = 25\left[\frac{1}{2}\left(\cosh^{-1}\left(\frac{x+5}{5}\right) - \frac{x+5}{5}\cdot\frac{\sqrt{(x+5)^2 - 25}}{5}\right) + C\right]$$ This is the indefinite integral for the given problem.