What is the integral of 1/(5+2t) ?

The integral of 1/(5+2t) is 1/2 ln|5+2t|+C

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integral of 1/(5+2t)

\int \frac{1}{5 + 2 t} d t

\frac{1}{2} ln ∣ 5 + 2 t ∣ + C

Solution steps

\int \frac{1}{5 + 2 t} d t

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 5 + 2 t

= \frac{1}{2} ln ∣ 5 + 2 t ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 5 + 2 t ∣ + C

\frac{\partial}{\partial a} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

2 (x - y) d x + d y = 0

x \to 1 + lim ((x)^{\frac{1}{1 - x}})

x + ∣ x ∣

((x + y)^{2}) d x + (2 x y + x^{2} - 8) d y = 0, y (1) = 1