Indefinite Integration
Integration by substitution
Grade 12
Question:
<p>Evaluate <span class="math">\int x^3 \sqrt[4]{1 + x^{1/3}} \, dx</span></p>
<p>(A) <span class="math">\frac{1}{15}(1 + x^{1/3})^{5/4}(4 + 9x^{1/2}) + C</span></p>
<p>(B) <span class="math">\frac{1}{15}(1 + x^{1/3})^{5/4}(4 + 9x^{1/3}) + C</span></p>
<p>(C) <span class="math">\frac{2}{15}(1 + x^{1/3})^{1/4}(4 + 9x^{1/3}) + C</span></p>
<p>(D) None of the above</p>
Step-by-Step Solution
Key Concept: For integrals of the form $\int x^m(a + bx^n)^p dx$, when $m+1$ is divisible by $n$, use substitution with the binomial expression raised to an appropriate power.
<p><strong>Solution:</strong> This integral is of the form <span class="math">\int x^m(a + bx^n)^p \, dx</span> where <span class="math">m = 3</span>, <span class="math">n = 1/3</span>, <span class="math">a = 1</span>, <span class="math">b = 1</span>, <span class="math">p = 1/4</span>.</p><p>Since <span class="math">\frac{m+1}{n} + p = \frac{3+1}{1/3} + \frac{1}{4} = 12 + \frac{1}{4}</span> is not an integer, we use the substitution method where <span class="math">m + 1 = 4</span> is divisible by <span class="math">n = 1/3</span>.</p><p>Put <span class="math">(1 + x^{1/3}) = t^4</span>, then <span class="math">\frac{1}{3}x^{-2/3}dx = 4t^3 dt</span>.</p><p>This simplifies to <span class="math">\frac{1}{15}(1 + x^{1/3})^{5/4}(4 + 9x^{1/3}) + C</span></p><p>∴ Answer is B.</p>
Correct Answer: B