How do you integrate #int (4x)/sqrt(3x)# using substitution?

Question

Samantha Adkison · Accepted Answer

I found: #8/27(3x)^(3/2)+c#

Have a look:

Scarlett Allison · Answer

#(8x^(3/2))/3^(3/2)+C#

However, we should see that substitution is a waste of time:

#int(4x)/sqrt(3x)dx=4/sqrt3intx/sqrtxdx=4/sqrt3intx^(1/2)dx#

Using the rule #intx^ndx=x^(n+1)/(n+1)+C#. where #n!=-1#, this becomes:

#=4/sqrt3(x^(3/2)/(3/2))+C=8/(3sqrt3)x^(3/2)+C=(8x^(3/2))/3^(3/2)+C#

Ezekiel Alexander · Answer

undefined To integrate $ \int \frac{4x}{\sqrt{3x}} $ using substitution, let's set $ u = \sqrt{3x} $. Then, we can find $ du $ in terms of $ dx $ and rewrite the integral in terms of $ u $. Here's how:

1. Let $ u = \sqrt{3x} $.
2. Find $ du $:
   $$ du = \frac{d}{dx}(\sqrt{3x})dx $$
   $$ du = \frac{1}{2\sqrt{3x}} \cdot 3dx $$
   $$ du = \frac{3}{2\sqrt{3x}}dx $$

3. Rewrite the integral in terms of $ u $:
   $$ \int \frac{4x}{\sqrt{3x}}dx = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{\sqrt{x}} dx $$
   $$ = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{\sqrt{3x}} \cdot \sqrt{3} dx $$
   $$ = \int \frac{4}{\sqrt{3}} \cdot \frac{x}{u} \cdot \sqrt{3} du $$
   $$ = 4\int \frac{1}{u} du $$
   $$ = 4 \ln|u| + C $$

4. Substitute back for $ u $:
   $$ = 4 \ln|\sqrt{3x}| + C $$
   $$ = 4 \ln|(\sqrt{3})(\sqrt{x})| + C $$
   $$ = 4 \ln(\sqrt{3}) + 4 \ln(\sqrt{x}) + C $$
   $$ = 2 \ln(3) + 2 \ln(x) + C $$

So, the integral of $ \frac{4x}{\sqrt{3x}} $ using substitution is $ 2 \ln(3) + 2 \ln(x) + C $, where $ C $ is the constant of integration.