What is the integral of 7/(4+7x) ?

The integral of 7/(4+7x) is ln|4+7x|+C

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integral of 7/(4+7x)

\int \frac{7}{4 + 7 x} d x

ln ∣ 4 + 7 x ∣ + C

Solution steps

\int \frac{7}{4 + 7 x} d x

Take the constant out:

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

= 7 \cdot \int \frac{1}{4 + 7 x} d x

Apply u-substitution

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

= 7 \cdot \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 7 \cdot \frac{1}{7} ln ∣ u ∣

Substitute back

u = 4 + 7 x

= 7 \cdot \frac{1}{7} ln ∣ 4 + 7 x ∣

Simplify

7 \cdot \frac{1}{7} ln ∣ 4 + 7 x ∣ : ln ∣ 4 + 7 x ∣

= ln ∣ 4 + 7 x ∣

Add a constant to the solution

= ln ∣ 4 + 7 x ∣ + C

\frac{\partial}{\partial x} (3 x^{2} y + x sin (2 y))

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

y^{^{'}} + sin (t) y = - sin (t), y (0) = 7

\frac{\partial}{\partial x} (\frac{1}{4} x^{a} y^{1 - a})

\int e^{- 2} d x