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

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

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

integral

\frac{t ^{2}}{( 1 - t ^{2} )}

\frac{ln ∣ t + 1 ∣}{2} - \frac{ln ∣ t - 1 ∣}{2} - t + C

Solution steps

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

\frac{t ^{2}}{1 - t ^{2}} = \frac{1}{1 - t ^{2}} - 1

= \int \frac{1}{1 - t ^{2}} - 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

= \int \frac{1}{1 - t ^{2}} d t - \int 1 d t

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

\int 1 d t = t

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

Add a constant to the solution

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

\frac{1}{( a - x )}

e^{\frac{- x ^{2}}{2}}

\frac{1}{x ^{n}}

\frac{8}{( x ^{3} - 16 x )}

ln (x + 2)

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