\( \int\displaystyle\frac{x^5}{\sqrt{1+x^3}}\,dx \) is equal to :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{2}{9}(1+x^3)^{3/2}-\displaystyle\frac{2}{3}(1+x^3)^{1/2}+C,\ where\ C\ is\ an\ arbitrary\ constant. \)
Let \( t=1+x^3 \). Then \( dt=3x^2\,dx \) and \( x^3=t-1 \), and writing \( x^5=x^3\cdot x^2 \):
\( \displaystyle \int\frac{x^3\cdot x^2\,dx}{\sqrt{1+x^3}}=\frac{1}{3}\int\frac{t-1}{\sqrt{t}}\,dt \)
\( \displaystyle =\frac{1}{3}\int\left(t^{1/2}-t^{-1/2}\right)dt \)
\( \displaystyle =\frac{1}{3}\left(\frac{2}{3}t^{3/2}-2t^{1/2}\right)+C \)
\( \displaystyle =\frac{2}{9}(1+x^3)^{3/2}-\frac{2}{3}(1+x^3)^{1/2}+C \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{2}{3}(1+x^3)^{3/2}+\displaystyle\frac{1}{3}(1+x^3)^{1/2}+C,\ where\ C\ is\ an\ arbitrary\ constant. \) — Incorrect. The coefficients should be \( \frac{2}{9} \) and \( -\frac{2}{3} \), not \( \frac{2}{3} \) and \( \frac{1}{3} \).
Option (2) \( \displaystyle\frac{2}{9}(1+x^3)^{1/2}-\displaystyle\frac{2}{3}(1+x^3)^{3/2}+C,\ where\ C\ is\ an\ arbitrary\ constant. \) — Incorrect. The coefficients are right but the two powers are interchanged.
Option (3) \( \displaystyle\frac{2}{3}(1+x^3)^{3/2}-\displaystyle\frac{1}{3}(1+x^3)^{1/2}+C,\ where\ C\ is\ an\ arbitrary\ constant. \) — Incorrect. The coefficients should be \( \frac{2}{9} \) and \( -\frac{2}{3} \), not \( \frac{2}{3} \) and \( -\frac{1}{3} \).