The value of \(\int\displaystyle\frac{x^5}{\sqrt{1+x^3}}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{2}{9}(1+x^3)^{3/2} - \frac{2}{3}(1+x^3)^{\frac{1}{2}} + c\) : \(c\) is an arbitrary constant
Write \(x^5 = x^3 \cdot x^2\) and use the substitution
\(t = 1 + x^3,\quad dt = 3x^2\,dx,\quad x^3 = t - 1\)
Then:
\( \displaystyle \int \frac{x^3 \cdot x^2}{\sqrt{1 + x^3}}\,dx = \frac{1}{3}\int \frac{t - 1}{\sqrt{t}}\,dt \)
\( \displaystyle = \frac{1}{3}\int \left(t^{1/2} - t^{-1/2}\right) dt \)
\( \displaystyle = \frac{1}{3}\left(\frac{2}{3}t^{3/2} - 2t^{1/2}\right) + c \)
\( \displaystyle = \frac{2}{9}(1 + x^3)^{3/2} - \frac{2}{3}(1 + x^3)^{1/2} + c \)