# How do i find integral of 7x+3/7x-4 dx?

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`int 7/(7x-4) dx`

I will just integrate this part of the problem.

There is a single power of x in the denominator, that suggests that we will have a log in our answer.

We will solve this using a u substitution.

Let u = 7x - 4

take the derivative of both sides to get:

`du = 7dx`

Now we can substitute u for 7x-4 and du for 7*dx

`int 1/u * du` which equals `ln u + c`

substitute 7x-4 back in for u

`ln(7x-4) + c`

Hope, this helps.

Find `int (7x+3)/(7x-4)dx `

Use long division on the integrand: `(7x+3)/(7x-4)=1+7/(7x-4) `

`int (7x+3)/(7x-4) dx `

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

`=int dx+int 7/(7x-4)dx ` (Use a u substitution: u=7x-4, du=7dx)

`=x+ln|7x-4|+C `

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`int (7x+3)/(7x-4)=x+ln|7x-4|+C `

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Let u=7x-4; then du=7dx.

Now `int 7/(7x-4)dx ` can be written as `int (du)/u ` .

Then the result is `ln|u|+C ` . Substitute u=7x+4 to get

`int (7dx)/(7x-4) = int (du)/u = ln|u|+C=ln|7x-4|+C `

Could you elaborate on the substitution part? I dont understand

I assume you mean integral of "`7x + 3/(7x) - 4` " (i.e., x is in denominator in the second term.

The integral can be done by remembering the basic integral formula.

`intx^n dx = x^(n+1) / (n+1)`

and, `intdx/x = ln x`

Therefore, we can calculate the given integral as:

`int (7x + 3/(7x) -4) dx = 7 int x dx + 3/7 int dx/x - 4 int dx = 7 x^2/2 +3/7 lnx - 4x +C`

where C is the constant of integration.

Hope this helps.

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`int` `(7x+3)/(7x-4)` dx

This is what i actually meant.