Question

Given \[ f(x)=\int \frac{dx}{x^{2/3}+2\sqrt{x}} \quad \text{and} \quad f(0)=-26+24\ln 2. \] If \(f(1)=A+B\ln 3\), then find \((A+B)\).

• A. \(10\)
• B. \(11\)
• C. \(-11\)
• D. \(-10\)

Hint:

When integrals involve mixed powers of \(x\), try a substitution that converts all powers into simple polynomials.

Answer 1

The Correct Option is C

Step 1: Simplify the integrand
\[ x^{2/3}+2\sqrt{x} = x^{1/2}\bigl(x^{1/6}+2\bigr) \] \[ \Rightarrow f(x)=\int \frac{dx}{x^{1/2}(x^{1/6}+2)} \] Let \[ x=t^6 \Rightarrow dx=6t^5dt \] Then: \[ x^{1/2}=t^3,\quad x^{1/6}=t \] \[ f(x)=\int \frac{6t^5dt}{t^3(t+2)} =6\int \frac{t^2}{t+2}\,dt \]
Step 2: Perform division
\[ \frac{t^2}{t+2}=t-2+\frac{4}{t+2} \] \[ f(x)=6\int\left(t-2+\frac{4}{t+2}\right)dt \] \[ =6\left(\frac{t^2}{2}-2t+4\ln|t+2|\right)+C \] \[ =3t^2-12t+24\ln|t+2|+C \] Substitute back \(t=x^{1/6}\): \[ f(x)=3x^{1/3}-12x^{1/6}+24\ln(x^{1/6}+2)+C \]
Step 3: Use the given condition \(f(0)\)
At \(x=0\): \[ f(0)=24\ln 2 + C \] Given: \[ f(0)=-26+24\ln 2 \Rightarrow C=-26 \]
Step 4: Find \(f(1)\)
\[ f(1)=3-12+24\ln 3-26 \] \[ f(1)=24\ln 3-35 \] Thus, \[ A=-35,\quad B=24 \]
Step 5: Compute \(A+B\)
\[ A+B=-35+24=-11 \] \[ \boxed{-11} \]