What is the integral of x/(7x-6) ?

The integral of x/(7x-6) is 1/49 (7x-6+6ln|7x-6|)+C

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integral of x/(7x-6)

\int \frac{x}{7 x - 6} d x

\frac{1}{49} (7 x - 6 + 6 ln ∣ 7 x - 6 ∣) + C

Solution steps

\int \frac{x}{7 x - 6} d x

Apply u-substitution

: \int \frac{u + 6}{49 u} d u

= \int \frac{u + 6}{49 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{49} \cdot \int \frac{u + 6}{u} d u

Expand

\frac{u + 6}{u} : 1 + \frac{6}{u}

= \frac{1}{49} \cdot \int 1 + \frac{6}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{49} (\int 1 d u + \int \frac{6}{u} d u)

\int 1 d u = u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

= \frac{1}{49} (u + 6 ln ∣ u ∣)

Substitute back

u = 7 x - 6

= \frac{1}{49} (7 x - 6 + 6 ln ∣ 7 x - 6 ∣)

Add a constant to the solution

= \frac{1}{49} (7 x - 6 + 6 ln ∣ 7 x - 6 ∣) + C

\frac{d}{d x} (1 - x^{4})

6 y^{^{''}} + y^{^{'}} - 2 y = 0

\frac{d ^{2}}{d x ^{2}} (e^{- k x^{2}})

\int 2 x 9 - x^{2} d x

x y^{^{'}} = y + 2 x^{2} y, y (1) = 1

What is the integral of x/(7x-6) ?
The integral of x/(7x-6) is 1/49 (7x-6+6ln|7x-6|)+C