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:
\( \alpha^4 = 64 \)
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)