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

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

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integral of (-1)/(y+1)

integral

\frac{- 1}{y + 1}

- ln ∣ y + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= - \int \frac{1}{y + 1} d y

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - ln ∣ u ∣

Substitute back

u = y + 1

= - ln ∣ y + 1 ∣

Add a constant to the solution

= - ln ∣ y + 1 ∣ + C

x^{2} e^{2} x

\frac{3}{x ^{4}}

\frac{e ^{x}}{x ^{4}}

\frac{tan ( x )}{1 + cos ^{4} ( x )}

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