What is the integral of 5/(2x+7) ?

The integral of 5/(2x+7) is 5/2 ln|2x+7|+C

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integral of 5/(2x+7)

\int \frac{5}{2 x + 7} d x

\frac{5}{2} ln ∣ 2 x + 7 ∣ + C

Solution steps

\int \frac{5}{2 x + 7} d x

Take the constant out:

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

= 5 \cdot \int \frac{1}{2 x + 7} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= 5 \cdot \int \frac{1}{2 u} d u

Take the constant out:

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

= 5 \cdot \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 5 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = 2 x + 7

= 5 \cdot \frac{1}{2} ln ∣ 2 x + 7 ∣

Simplify

5 \cdot \frac{1}{2} ln ∣ 2 x + 7 ∣ : \frac{5}{2} ln ∣ 2 x + 7 ∣

= \frac{5}{2} ln ∣ 2 x + 7 ∣

Add a constant to the solution

= \frac{5}{2} ln ∣ 2 x + 7 ∣ + C

x \to 2 lim (\frac{x + 4}{3 x - 1})

\int \frac{1}{( 4 x ^{2} - 1 ) ^{2}} d x

\int 2 x (x^{2} + 8)^{7} d x

\frac{\partial}{\partial x} ((y - x^{2})^{2} + x^{5})

g (x) = \frac{1}{( 1 + 2 x ) ^{3}}