What is the integral of 1/(7x+8) ?

The integral of 1/(7x+8) is 1/7 ln|7x+8|+C

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integral of 1/(7x+8)

\int \frac{1}{7 x + 8} d x

\frac{1}{7} ln ∣ 7 x + 8 ∣ + C

Solution steps

\int \frac{1}{7 x + 8} d x

Apply u-substitution

: \int \frac{1}{7 u} d u

= \int \frac{1}{7 u} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{7} ln ∣ u ∣

Substitute back

u = 7 x + 8

= \frac{1}{7} ln ∣ 7 x + 8 ∣

Add a constant to the solution

= \frac{1}{7} ln ∣ 7 x + 8 ∣ + C

(0.1 x^{2} - 4.2 x + 6.5)^{1.5}

e^{s e c (8 x)}

n = 1 \sum \infty \frac{5}{2 n ^{2} + 1}

\int \frac{1}{1} (2 x) d x

\frac{\partial}{\partial x} (x^{2} - 2 y^{2} + 4)