What is the integral of 2/(t-t^3) ?

The integral of 2/(t-t^3) is 2ln|t|-ln|t+1|-ln|t-1|+C

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integral of 2/(t-t^3)

\int \frac{2}{t - t ^{3}} d t

2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ + C

Solution steps

\int \frac{2}{t - t ^{3}} 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 - t ^{3}} d t

Take the partial fraction of

\frac{1}{t - t ^{3}} : \frac{1}{t} - \frac{1}{2 ( t + 1 )} - \frac{1}{2 ( t - 1 )}

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{t} d t = ln ∣ t ∣

\int \frac{1}{2 ( t + 1 )} d t = \frac{1}{2} ln ∣ t + 1 ∣

\int \frac{1}{2 ( t - 1 )} d t = \frac{1}{2} ln ∣ t - 1 ∣

= 2 (ln ∣ t ∣ - \frac{1}{2} ln ∣ t + 1 ∣ - \frac{1}{2} ln ∣ t - 1 ∣)

Simplify

2 (ln ∣ t ∣ - \frac{1}{2} ln ∣ t + 1 ∣ - \frac{1}{2} ln ∣ t - 1 ∣) : 2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣

= 2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣

Add a constant to the solution

= 2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ + C

\frac{d y}{d x} = \frac{9 x + 2 y}{9 x + 2 y + 2}

\frac{\partial}{\partial x} (x \frac{e ^{y}}{x + y})

x \to 0 lim (ln (x))

\int 29 arctan (x) d x

What is the integral of 2/(t-t^3) ?
The integral of 2/(t-t^3) is 2ln|t|-ln|t+1|-ln|t-1|+C