Question

$\int \frac{(x^4 - x)^{1/4}}{x^5} dx$ is equal to

• A. $\frac{4}{15}(1-\frac{1}{x^3})^{5/4}+c$
• B. $\frac{4}{5}(1-\frac{1}{x^3})^{5/4}+c$
• C. $\frac{4}{15}(1+\frac{1}{x^3})^{5/4}+c$
• D. None of these

Answer 1

Answer: (A)

$\frac{4}{15}(1-\frac{1}{x^3})^{5/4}+c$

Answer 2

The integral to evaluate is $I = \int \frac{(x^4 - x)^{1/4}}{x^5} dx$.

We can factor out $x^4$ from the term inside the parenthesis in the numerator: $(x^4 - x)^{1/4} = (x^4(1 - \frac{x}{x^4}))^{1/4} = (x^4(1 - \frac{1}{x^3}))^{1/4}$.

Using the property $(ab)^n = a^n b^n$, we get: $(x^4(1 - \frac{1}{x^3}))^{1/4} = (x^4)^{1/4} (1 - \frac{1}{x^3})^{1/4}$.

For real numbers, $(x^4)^{1/4} = |x|$.

So the integral becomes $I = \int \frac{|x| (1 - \frac{1}{x^3})^{1/4}}{x^5} dx$.

For the expression $(x^4 - x)^{1/4}$ to be a real number, we must have $x^4 - x \ge 0$. $x^4 - x = x(x^3 - 1) = x(x-1)(x^2+x+1)$. The quadratic factor $x^2+x+1 = (x + 1/2)^2 + 3/4$ is always positive. So $x(x-1)(x^2+x+1) \ge 0$ implies $x(x-1) \ge 0$. This inequality holds for $x \le 0$ or $x \ge 1$.

Let's consider the case where $x \ge 1$. In this case, $x > 0$, so $|x| = x$. The integral becomes $I = \int \frac{x (1 - \frac{1}{x^3})^{1/4}}{x^5} dx = \int \frac{(1 - \frac{1}{x^3})^{1/4}}{x^4} dx$. Let $u = 1 - \frac{1}{x^3}$. Then $du = \frac{d}{dx}(1 - x^{-3}) dx = (0 - (-3)x^{-4}) dx = 3x^{-4} dx = \frac{3}{x^4} dx$. So, $\frac{1}{x^4} dx = \frac{1}{3} du$. Substituting this into the integral: $I = \int u^{1/4} \cdot \frac{1}{3} du = \frac{1}{3} \int u^{1/4} du$. Using the power rule for integration, $\int u^n du = \frac{u^{n+1}}{n+1} + C$ (for $n \neq -1$): $I = \frac{1}{3} \frac{u^{1/4 + 1}}{1/4 + 1} + C = \frac{1}{3} \cdot \frac{u^{5/4}}{5/4} + C$. $I = \frac{1}{3} \cdot \frac{4}{5} u^{5/4} + C = \frac{4}{15} u^{5/4} + C$. Substitute back $u = 1 - \frac{1}{x^3}$: $I = \frac{4}{15} (1 - \frac{1}{x^3})^{5/4} + C$. This result is valid for $x \ge 1$.