Evaluate $\int \frac{x^{1/2}}{1 + x^{3/4}} \, dx$.

$\frac{4}{3} \left[ x^{3/4} - \ln(1 + x^{3/4}) \right] + C$

The correct answer is Option (1) → $\frac{4}{3} \left[ x^{3/4} - \ln(1 + x^{3/4}) \right] + C$

$I = \int \frac{x^{1/2}}{1 + x^{3/4}} \, dx \text{}$

Put $x = t^4 ⇒dx = 4t^3 \, dt$

$∴I = 4 \int \frac{t^2(t^3)}{1 + t^3} \, dt = 4 \int \left( t^2 - \frac{t^2}{1 + t^3} \right) dt$ $[∵x^{1/2} = t^2]$

$⇒I = 4 \int t^2 \, dt - 4 \int \frac{t^2}{1 + t^3} \, dt$

Let $I = I_1 - I_2$

Now, $I_1 = 4 \int t^2 \, dt = 4 \cdot \frac{t^3}{3} + C_1 = \frac{4}{3}x^{3/4} + C_1$

and $I_2 = 4 \int \frac{t^2}{1 + t^3} \, dt$

Again, put $1 + t^3 = z ⇒ 3t^2 \, dt = dz$

$⇒t^2 \, dt = \frac{1}{3} \, dz$

Now, $I_2 = \frac{4}{3} \int \frac{1}{z} \, dz$

$= \frac{4}{3} \log |z| + C_2 = \frac{4}{3} \log |(1 + t^3)| + C_2$

$= \frac{4}{3} \log (1 + x^{3/4}) + C_2$

$∴I = \frac{4}{3}x^{3/4} + C_1 - \frac{4}{3} \log |(1 + x^{3/4})| - C_2$

$= \frac{4}{3} \left[ x^{3/4} - \log |(1 + x^{3/4})| \right] + C$ $[∵C = C_1 - C_2]$