What is the integral of (2-4t^2)/(t-t^3) ?

The integral of (2-4t^2)/(t-t^3) is 2ln|t|+ln|t+1|+ln|t-1|+C

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integral of (2-4t^2)/(t-t^3)

\int \frac{2 - 4 t ^{2}}{t - t ^{3}} d t

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

Solution steps

\int \frac{2 - 4 t ^{2}}{t - t ^{3}} d t

Expand

\frac{2 - 4 t ^{2}}{t - t ^{3}} : \frac{2}{t - t ^{3}} - \frac{4 t ^{2}}{t - t ^{3}}

Apply the Sum Rule:

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

= \int \frac{2}{t - t ^{3}} d t - \int \frac{4 t ^{2}}{t - t ^{3}} d t

\int \frac{2}{t - t ^{3}} d t = 2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣

\int \frac{4 t ^{2}}{t - t ^{3}} d t = 4 (- \frac{1}{2} ln ∣ t + 1 ∣ - \frac{1}{2} ln ∣ t - 1 ∣)

= 2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ - 4 (- \frac{1}{2} ln ∣ t + 1 ∣ - \frac{1}{2} ln ∣ t - 1 ∣)

Simplify

2 ln ∣ t ∣ - ln ∣ t + 1 ∣ - ln ∣ t - 1 ∣ - 4 (- \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}{d x} ((x^{2})^{2 x})

\int_{0}^{\frac{π}{6}} 17 sec^{4} (x) d x

\frac{8 s ^{2} - 4 s + 12}{s ( s ^{2} + 4 )}

f (x) = x, at x = \frac{49}{121}

\frac{d}{d x} (\frac{1}{2 - x})

What is the integral of (2-4t^2)/(t-t^3) ?
The integral of (2-4t^2)/(t-t^3) is 2ln|t|+ln|t+1|+ln|t-1|+C