If int frac{dx}{x^{7/2}(x^4+1)^{3/8}} = m lef | vedclass

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Question

(A) Let $I = \int \frac{dx}{x^{7/2}(x^4+1)^{3/8}}$.Factor out $x^4$ from the term in the parenthesis: $I = \int \frac{dx}{x^{7/2} [x^4(1 + x^{-4})]^{3

Vedclass · Accepted Answer

$-\frac{1}{16}$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{x^{7/2}(x^4+1)^{3/8}}$.Factor out $x^4$ from the term in the parenthesis: $I = \int \frac{dx}{x^{7/2} [x^4(1 + x^{-4})]^{3/8}}$.$I = \int \frac{dx}{x^{7/2} \cdot x^{4 \cdot (3/8)} (1 + x^{-4})^{3/8}} = \int \frac{dx}{x^{7/2} \cdot x^{3/2} (1 + x^{-4})^{3/8}}$.$I = \int \frac{dx}{x^{10/2} (1 + x^{-4})^{3/8}} = \int \frac{dx}{x^5 (1 + x^{-4})^{3/8}}$.Let $u = 1 + x^{-4}$, then $du = -4x^{-5} dx$, which implies $x^{-5} dx = -\frac{1}{4} du$.Substituting these into the integral: $I = \int (u)^{-3/8} \cdot (-\frac{1}{4} du) = -\frac{1}{4} \int u^{-3/8} du$.$I = -\frac{1}{4} \cdot \frac{u^{(-3/8 + 1)}}{(-3/8 + 1)} + c = -\frac{1}{4} \cdot \frac{u^{5/8}}{5/8} + c = -\frac{1}{4} \cdot \frac{8}{5} u^{5/8} + c = -\frac{2}{5} (1 + x^{-4})^{5/8} + c$.Rewriting the expression: $I = -\frac{2}{5} \left(\frac{x^4+1}{x^4}\right)^{5/8} + c$.Comparing this with $m \left(\frac{x^4+1}{x^4}\right)^n + c$, we get $n = \frac{5}{8}$. Since the options provided do not match $\frac{5}{8}$, we re-evaluate the integral form. Given the structure, the correct value of $n$ is $5/8$.

If $\int \frac{dx}{x^{7/2}(x^4+1)^{3/8}} = m \left(\frac{x^4+1}{x^4}\right)^n + c$, where $c$ is a constant of integration, then the value of $n$ is...

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