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

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

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integral of 1/(1+2y)

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

\frac{1}{2} ln ∣ 1 + 2 y ∣ + C

Solution steps

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

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 = 1 + 2 y

= \frac{1}{2} ln ∣ 1 + 2 y ∣

Add a constant to the solution

= \frac{1}{2} ln ∣ 1 + 2 y ∣ + C

\int \frac{1}{x 3 - x ^{2}} d x

d y = x^{2} d x, y (0) = 0

\int \frac{x}{1 - x ^{3}} d x

\int \frac{sin ( x )}{( 1 + cos ( x ) )} d x

\frac{1}{x + 8}