Question

Let I(x) = \(∫√\frac{x+7}{x} dx\) and I (9) = 12 + 7 loge 7. If I (1) = α + 7 loge (1 + 2√2), then α4 is equal to_____.

Answer 1

Correct Answer: 64

The given integral is:

\( \int \sqrt{\frac{x+7}{x}} \, dx \)

Step 1: Substitution

Let \( x = t^2 \), so:

\( dx = 2t \, dt \)

Substitute into the integral:

\( \int \sqrt{\frac{x+7}{x}} \, dx = 2 \int \sqrt{\frac{t^2+7}{t^2}} t \, dt \)

Simplify:

\( 2 \int \sqrt{t^2 + 7} \, dt = 2 \int (t + \sqrt{7}) \, dt \)

Integrate:

\( I(t) = 2 \left[ \frac{t}{2} \sqrt{t^2 + 7} + \frac{7}{2} \ln |t + \sqrt{t^2 + 7}| \right] + C \)

Simplify:

\( I(t) = t\sqrt{t^2 + 7} + 7 \ln |t + \sqrt{t^2 + 7}| + C \)

Substitute \( t = \sqrt{x} \) back:

\( I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln |\sqrt{x} + \sqrt{x+7}| + C \)

Step 2: Finding the constant \( C \)

Given \( I(9) = 12 + 7 \ln 7 \):

\( I(9) = \sqrt{9} \sqrt{9+7} + 7 \ln |\sqrt{9} + \sqrt{9+7}| + C \)

Simplify:

\( 12 + 7 \ln 7 = 12 + 7 \ln (3+4) + C \)

\( C = 0 \)

Thus:

\( I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln |\sqrt{x} + \sqrt{x+7}| \)

Step 3: Finding \( I(1) \) and \( \alpha \)

Substitute \( x=1 \):

\( I(1) = \sqrt{1} \sqrt{1+7} + 7 \ln |\sqrt{1} + \sqrt{1+7}| \)

\( I(1) = \sqrt{8} + 7\ln(1 + \sqrt{8}) \)

Let \( \alpha = \sqrt{8} \), so:

\( \alpha^4 = (\sqrt{8})^4 = 8^2 = 64 \)

Final Answer:

\( \alpha^4 = 64 \)