Question

this is similar to section 4.2 problem 22:
determine the indefinite integral. use capital c for the free
$$ int \frac { 6 } { sqrt 4 { x ^ { 5 } } } d x = $$
hint: follow example 6.
resources
ebook

Explanation:

Step1: Rewrite the integrand

Rewrite \(\frac{6}{\sqrt[4]{x^{5}}}\) as \(6x^{-\frac{5}{4}}\) using the rule \(\frac{1}{x^{n}}=x^{-n}\) and \(\sqrt[m]{x^{n}} = x^{\frac{n}{m}}\).

Step2: Apply the power rule for integration

The power rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int6x^{-\frac{5}{4}}dx\), since \(\int kf(x)dx=k\int f(x)dx\) (\(k = 6\) is a constant), we have \(6\int x^{-\frac{5}{4}}dx\).
Using the power rule with \(n=-\frac{5}{4}\), we get \(6\times\frac{x^{-\frac{5}{4}+1}}{-\frac{5}{4}+1}+C\).
Simplify the exponent and the denominator: \(-\frac{5}{4}+1=-\frac{1}{4}\).
So, \(6\times\frac{x^{-\frac{1}{4}}}{-\frac{1}{4}}+C\).

Step3: Simplify the expression

\(6\times\frac{x^{-\frac{1}{4}}}{-\frac{1}{4}}+C=6\times(- 4)x^{-\frac{1}{4}}+C=-24x^{-\frac{1}{4}}+C\).
We can also rewrite \(x^{-\frac{1}{4}}\) as \(\frac{1}{\sqrt[4]{x}}\), so the integral is \(-\frac{24}{\sqrt[4]{x}}+C\).

Answer:

\(-\frac{24}{\sqrt[4]{x}}+C\)