Question

If \(I=\int \frac{x^5}{\sqrt{1+x^3}}\,dx\), then \(I\) is equal to

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

Hint:

When integrand has \(\sqrt{1+x^3}\), use substitution \(t=1+x^3\). Always rewrite higher powers in terms of \(t\).

Answer 1

The Correct Option is D

Step 1: Substitute \(t = 1+x^3\).
\[ t = 1+x^3 \Rightarrow dt = 3x^2 dx \Rightarrow x^2 dx = \frac{dt}{3} \]
Step 2: Rewrite the integrand.
\[ I=\int \frac{x^5}{\sqrt{1+x^3}}dx = \int \frac{x^3\cdot x^2}{\sqrt{t}}dx \]
But \(x^3 = t-1\). So:
\[ I=\int \frac{(t-1)\cdot x^2}{\sqrt{t}}dx = \int \frac{(t-1)}{\sqrt{t}} \cdot \frac{dt}{3} \]
Step 3: Simplify.
\[ I=\frac{1}{3}\int \left(t^{\frac{1}{2}} - t^{-\frac{1}{2}}\right) dt \]
Step 4: Integrate term by term.
\[ \int t^{\frac{1}{2}}dt = \frac{2}{3}t^{\frac{3}{2}} \] \[ \int t^{-\frac{1}{2}}dt = 2t^{\frac{1}{2}} \]
So:
\[ I=\frac{1}{3}\left(\frac{2}{3}t^{\frac{3}{2}} - 2t^{\frac{1}{2}}\right)+C \]
\[ I=\frac{2}{9}t^{\frac{3}{2}}-\frac{2}{3}t^{\frac{1}{2}}+C \]
Step 5: Substitute back \(t=1+x^3\).
\[ I=\frac{2}{9}(1+x^3)^{\frac{3}{2}}-\frac{2}{3}(1+x^3)^{\frac{1}{2}}+C \]
Final Answer:
\[ \boxed{\frac{2}{9}(1+x^3)^{\frac{3}{2}}-\frac{2}{3}(1+x^3)^{\frac{1}{2}}+C} \]