Integration prob: int cube root of 5x dx by substitution

[mooshupork34]

I have to integrate the cube root of 5x dx by substitution.

so far, i have equated the cube root of 5x dx with (5x)^(1/3)

after this, I'm not really sure where to go so any i would be grateful for any help!!

 

[galactus]

Try letting \(\displaystyle \L\\u=5x, \;\ \frac{1}{5}du=dx\)

 

[mooshupork34]

Okay, so doing that, I got

the integral of u^1/3 * 1/5 du

or 5x^1/3 * 1/5 du

I tried to integrate that using substitution, and I set u = 5x^1/3. For du/dx, I got 5/3x^2/3 . This meant that dx = (3x^2/3) /5 du.

After this I got the integral of 1/5 * u * (3x^2/3) /5 du.

I'm not really sure where to go from there.

 

[skeeter]

\(\displaystyle \L u = 5x\)
\(\displaystyle \L du = 5 dx\)

\(\displaystyle \L \int (5x)^{\frac{1}{3}} dx\)

\(\displaystyle \L \frac{1}{5} \L \int u^{\frac{1}{3}} du\)

\(\displaystyle \L\frac{1}{5} \cdot \frac{3}{4}u^{\frac{4}{3}} + C\)

\(\displaystyle \L \frac{3}{20}(5x)^{\frac{4}{3}} + C\)