Question

If I = $\int_{}^{}\frac{x^{5}}{\sqrt{1 + x^{3}}}$dx, then I is equal to:

• A. $\frac{2}{9}$ (1 + x³)^5/2 + $\frac{2}{3}$ (1+ x³)^3/2 + C
• B. $\frac{2}{9}$ (1 + x³)^3/2–$\frac{2}{3}$ (1 + x³)^1/2 + C
• C. log$\left| \sqrt{x} + \sqrt{1 + x^{3}} \right|$+ C
• D. x² log (1 + x³) + C

Answer 1

Answer: (B)

$\frac{2}{9}$ (1 + x³)^3/2–$\frac{2}{3}$ (1 + x³)^1/2 + C

Answer 2

Put 1 + x³ = t², 3x² dx = 2t dt, so that

I = . $\frac{2}{3}$t dt = $\frac{2}{3}$ $\left( \frac{t^{3}}{3} - t \right)$+ C

= $\frac{2}{9}$ (1+ x³)^3/2 – $\frac{2}{3}$ (1 + x³)^1/2 + C