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consider the indefinite integral \\(\\int x^4 \\cdot \\sqrt2{x^5 - 8} \…

Question

consider the indefinite integral \\(\int x^4 \cdot \sqrt2{x^5 - 8} \\, dx\\):

this can be transformed into a basic integral by letting

\\(u = \square\\) and

\\(du = \square dx\\)

performing the substitution yields the integral

\\(\int \square \\, du\\)

Explanation:

Step1: Choose substitution variable

Let $u = x^5 - 8$

Step2: Compute derivative of u

$\frac{du}{dx} = 5x^4 \implies du = 5x^4 dx$

Step3: Rewrite integral in terms of u

From $du = 5x^4 dx$, we get $x^4 dx = \frac{1}{5}du$. Substitute into original integral:
$\int \sqrt[2]{u} \cdot \frac{1}{5} du = \int \frac{1}{5}u^{\frac{1}{2}} du$

Answer:

$u = x^5 - 8$
$du = 5x^4$
$\int \frac{1}{5}u^{\frac{1}{2}}$