What is the integral of x/(6x-5) ?

The integral of x/(6x-5) is 1/36 (6x-5+5ln|6x-5|)+C

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integral of x/(6x-5)

\int \frac{x}{6 x - 5} d x

\frac{1}{36} (6 x - 5 + 5 ln ∣ 6 x - 5 ∣) + C

Solution steps

\int \frac{x}{6 x - 5} d x

Apply u-substitution

: \int \frac{u + 5}{36 u} d u

= \int \frac{u + 5}{36 u} d u

Take the constant out:

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

= \frac{1}{36} \cdot \int \frac{u + 5}{u} d u

Expand

\frac{u + 5}{u} : 1 + \frac{5}{u}

= \frac{1}{36} \cdot \int 1 + \frac{5}{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}{36} (\int 1 d u + \int \frac{5}{u} d u)

\int 1 d u = u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{1}{36} (u + 5 ln ∣ u ∣)

Substitute back

u = 6 x - 5

= \frac{1}{36} (6 x - 5 + 5 ln ∣ 6 x - 5 ∣)

Add a constant to the solution

= \frac{1}{36} (6 x - 5 + 5 ln ∣ 6 x - 5 ∣) + C

y^{^{'}} = 3 y^{\frac{2}{3}}

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

f (x) = sin^{4} (x)

\frac{d}{d x} (100 x^{2})

y^{^{'}} = (x + y - 2) 2

What is the integral of x/(6x-5) ?
The integral of x/(6x-5) is 1/36 (6x-5+5ln|6x-5|)+C