Question

find the integral.

int \frac { 2 } { sqrt 4 { x ^ { 3 } } } d x

int \frac { 2 } { sqrt 4 { x ^ { 3 } } } d x =

Explanation:

Step1: Simplify the integrand

First, rewrite the integrand \(\frac{2}{\sqrt[4]{x^{3}}}\). Using the rule \(x^{m/n}=\sqrt[n]{x^{m}}\) and \(\frac{1}{x^{k}} = x^{-k}\), we have \(\frac{2}{\sqrt[4]{x^{3}}}=2x^{-\frac{3}{4}}\).

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 \(\int2x^{-\frac{3}{4}}dx\), since \(\int kf(x)dx=k\int f(x)dx\) (\(k = 2\) is a constant), then \(\int2x^{-\frac{3}{4}}dx=2\int x^{-\frac{3}{4}}dx\).
Substituting \(n=-\frac{3}{4}\) into the power - rule formula: \(2\times\frac{x^{-\frac{3}{4}+1}}{-\frac{3}{4}+1}+C\).
Calculate \(-\frac{3}{4}+1=\frac{1}{4}\).
So \(2\times\frac{x^{\frac{1}{4}}}{\frac{1}{4}}+C\).
Simplify \(2\times4x^{\frac{1}{4}}+C = 8\sqrt[4]{x}+C\).

Answer:

\(8\sqrt[4]{x}+C\)