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

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

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integral of (-2)/(t+1)

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

- 2 ln ∣ t + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= - 2 \cdot \int \frac{1}{t + 1} d t

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - 2 ln ∣ u ∣

Substitute back

u = t + 1

= - 2 ln ∣ t + 1 ∣

Add a constant to the solution

= - 2 ln ∣ t + 1 ∣ + C

\frac{1200}{s ^{2} ( s + 10 ) ^{2}}

\frac{2 x ^{3} - 6}{x ^{4}}

\frac{d}{d x} (\frac{x cos ( x ) - sin ( x )}{x ^{2}})

\frac{\partial}{\partial x} (30 x^{5} y^{4} - 20 x^{4} y^{3})

x \to - 2 lim (\frac{1}{x ^{2}})