What is the integral of 6/(6x+1) ?

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

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integral of 6/(6x+1)

\int \frac{6}{6 x + 1} d x

ln ∣ 6 x + 1 ∣ + C

Solution steps

\int \frac{6}{6 x + 1} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{6 x + 1} d x

Apply u-substitution

: \int \frac{1}{6 u} d u

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 6 \cdot \frac{1}{6} ln ∣ u ∣

Substitute back

u = 6 x + 1

= 6 \cdot \frac{1}{6} ln ∣ 6 x + 1 ∣

Simplify

6 \cdot \frac{1}{6} ln ∣ 6 x + 1 ∣ : ln ∣ 6 x + 1 ∣

= ln ∣ 6 x + 1 ∣

Add a constant to the solution

= ln ∣ 6 x + 1 ∣ + C

f (x) = sec (x), at x = - \frac{7 π}{4}

\int (x^{2} + 1)^{3} x d x

\frac{d}{d x} (\frac{4 x ^{2} + 6 x + 8}{x})

\int \frac{y}{1 + x ^{2}} d y

f (x) = 4 + 4 x^{2} - 2 x^{3}, at x = a