Questions: Find the indefinite integral by making a change of variables. (Hint: Let u be the denominator of the integrand. Remember to use absolute values where appropriate. Remember the constant of integration.) ∫ sqrt(x)/(sqrt(x)-6) dx

Solution Steps

To solve the integral \(\int \frac{\sqrt{x}}{\sqrt{x}-6} \, dx\), we will use the substitution method. The hint suggests letting \(u\) be the denominator of the integrand. Therefore, we choose: \[ u = \sqrt{x} - 6 \]

From the substitution \(u = \sqrt{x} - 6\), we can express \(\sqrt{x}\) as: \[ \sqrt{x} = u + 6 \]

Squaring both sides, we get: \[ x = (u + 6)^2 \]

Differentiate \(x = (u + 6)^2\) with respect to \(u\): \[ \frac{dx}{du} = 2(u + 6) \]

Thus, \(dx\) can be expressed as: \[ dx = 2(u + 6) \, du \]

Substitute \(u = \sqrt{x} - 6\), \(\sqrt{x} = u + 6\), and \(dx = 2(u + 6) \, du\) into the integral: \[ \int \frac{\sqrt{x}}{\sqrt{x}-6} \, dx = \int \frac{u + 6}{u} \cdot 2(u + 6) \, du \]

Simplify the integrand: \[ = \int \frac{(u + 6) \cdot 2(u + 6)}{u} \, du = \int \frac{2(u + 6)^2}{u} \, du \]

Expand \((u + 6)^2\): \[ (u + 6)^2 = u^2 + 12u + 36 \]

Substitute back into the integral: \[ = \int \frac{2(u^2 + 12u + 36)}{u} \, du \]

Simplify the expression: \[ = \int \left(2u + 24 + \frac{72}{u}\right) \, du \]

Integrate each term separately: \[ \int 2u \, du = u^2 \] \[ \int 24 \, du = 24u \] \[ \int \frac{72}{u} \, du = 72 \ln |u| \]

Combine the results: \[ = u^2 + 24u + 72 \ln |u| + C \]

Recall that \(u = \sqrt{x} - 6\). Substitute back: \[ = (\sqrt{x} - 6)^2 + 24(\sqrt{x} - 6) + 72 \ln |\sqrt{x} - 6| + C \]

Final Answer