Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>[JEE Main 2021] \(\displaystyle\int x^5\sqrt{1+x^3}\,dx\) equals (where \(C\) is a constant)</p>
<li>\(\dfrac{2}{9}(1+x^3)^{3/2}(3x^3-2)+C\)</li>
<li>\(\dfrac{2}{15}(1+x^3)^{5/2}-\dfrac{2}{9}(1+x^3)^{3/2}+C\)</li>
<li>\(\dfrac{2}{9}(1+x^3)^{3/2}-\dfrac{2}{15}(1+x^3)^{5/2}+C\)</li>
<li>\(\dfrac{1}{5}x^5\sqrt{1+x^3}+C\)</li>
Step-by-Step Solution
Key Concept: Substitute t=1+x^3, dt=3x^2dx. Express x^5=(x^3) \cdot x^2 = (t-1) \cdot (dt/3). Then \int(t-1)\sqrt{t} \cdot (dt/3).
<p><strong>Substitution:</strong> $t=1+x^3\Rightarrow dt=3x^2\,dx,\;x^3=t-1$.</p>
<p>$$x^5\sqrt{1+x^3}\,dx = x^3\cdot x^2\sqrt{1+x^3}\,dx = (t-1)\sqrt{t}\cdot\frac{dt}{3}$$</p>
<p>$$= \frac13\int(t^{3/2}-t^{1/2})\,dt = \frac13\left[\frac{2t^{5/2}}{5}-\frac{2t^{3/2}}{3}\right]+C$$</p>
<p>$$= \frac{2t^{5/2}}{15}-\frac{2t^{3/2}}{9}+C = \frac{2}{15}(1+x^3)^{5/2}-\frac{2}{9}(1+x^3)^{3/2}+C$$</p>
<p>Answer: <strong>(B)</strong></p>
Correct Answer: B