How do you find the integral of #(4x)/(4x+7)dx#?

Question

Ezra Adams · Accepted Answer

I like rewriting: #(4x)/(4x+7) = (4x+7-7)/(4x+7) = 1- 7/(4x+7)#

#int (4x)/(4x+7) = int( 1- 7/(4x+7)) dx#

#= x-7/4 ln abs(4x+7) +C#

Note To evaluate #int 7/(4x+7) dx# use substitution with #u= 4x+7# so #dx = 1/4 du# and we have #7/4 int 1/u du#

Second Note

This integral can also be evaluated by integration by parts, with #u=x# and #dv = 4/(4x+7)#.

Parts gives us:

#xln abs(4x+7) - int ln (4x+7) dx#

The integral here can be found by substitution if you know #int lnu du# and by substitution and integration by parts if you don't know #int ln u du#.

Emily Ammerman · Answer

undefined To find the integral of $ \frac{4x}{4x+7} $ with respect to $ x $, you can use the method of substitution. Let $ u = 4x + 7 $. Then, $ du = 4dx $. Rearrange this to solve for $ dx $ and substitute into the integral. You will get:

$$ \int \frac{4x}{4x+7} \, dx = \int \frac{1}{4} \frac{4}{u} \, du $$

The $ 4 $ in the numerator and the denominator cancel out, leaving you with:

$$ \frac{1}{4} \int \frac{1}{u} \, du $$

This integral is straightforward to evaluate:

$$ \frac{1}{4} \ln|u| + C $$

Substitute back $ u = 4x + 7 $:

$$ \frac{1}{4} \ln|4x + 7| + C $$

So, the integral of $ \frac{4x}{4x+7} $ with respect to $ x $ is $ \frac{1}{4} \ln|4x + 7| + C $, where $ C $ is the constant of integration.