Questions: Find the indefinite integral by making a change of variables. ∫ (sqrt(x))/(sqrt(x)-1) dx

Solution Steps

To solve the integral \(\int \frac{\sqrt{x}}{\sqrt{x}-1} \, dx\), we can use a substitution method. Let's set \( u = \sqrt{x} - 1 \), which implies \( \sqrt{x} = u + 1 \). Then, differentiate both sides with respect to \( x \) to find \( dx \) in terms of \( du \). Substitute these expressions into the integral and simplify to find the integral in terms of \( u \).

We start with the integral

\[ \int \frac{\sqrt{x}}{\sqrt{x}-1} \, dx. \]

We make the substitution \( u = \sqrt{x} - 1 \), which gives us \( \sqrt{x} = u + 1 \). Consequently, we can express \( x \) in terms of \( u \) as follows:

\[ x = (u + 1)^2. \]

Next, we differentiate \( x \) with respect to \( u \):

\[ dx = 2(u + 1) \, du. \]

Now we substitute \( \sqrt{x} \) and \( dx \) into the integral:

\[ \int \frac{(u + 1)}{u} \cdot 2(u + 1) \, du = \int 2(u + 1)^2 \cdot \frac{1}{u} \, du. \]

This simplifies to:

\[ \int 2 \left( u + 1 \right)^2 \cdot \frac{1}{u} \, du. \]

The integral can be expanded and integrated:

\[ \int 2 \left( u^2 + 2u + 1 \right) \cdot \frac{1}{u} \, du = \int 2 \left( u + 2 + \frac{1}{u} \right) \, du. \]

This results in:

\[ 2 \left( \frac{u^2}{2} + 2u + \log |u| \right) = u^2 + 4u + 2\log |u|. \]

Finally, we substitute back \( u = \sqrt{x} - 1 \):

\[ (\sqrt{x} - 1)^2 + 4(\sqrt{x} - 1) + 2\log |\sqrt{x} - 1|. \]

Final Answer