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

9 x - 4 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = 2

f (x) = 5 - 2 x^{2}, at x = - 2

\int \frac{2 x + 1}{( - 4 x ^{2} + 4 x + 3 ) ^{\frac{3}{2}}} d x

\frac{d}{d x} (sin (3 x + 2))

\frac{4}{( s + 10 ) ( s ^{2} + 16 ) ^{2}}