What is the integral of t/(1-t^8) ?

The integral of t/(1-t^8) is 1/2 (1/2 arctan(t^2)+1/4 ln|t^2+1|-1/4 ln|t^2-1|)+C

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integral of t/(1-t^8)

\int \frac{t}{1 - t ^{8}} d t

\frac{1}{2} (\frac{1}{2} arctan (t^{2}) + \frac{1}{4} ln t^{2} + 1 - \frac{1}{4} ln t^{2} - 1) + C

Solution steps

\int \frac{t}{1 - t ^{8}} d t

Apply u-substitution

: \int \frac{1}{2 ( 1 - u ^{4} )} d u

= \int \frac{1}{2 ( 1 - u ^{4} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{1 - u ^{4}} d u

Take the partial fraction of

\frac{1}{1 - u ^{4}} : \frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )}

= \frac{1}{2} \cdot \int \frac{1}{2 ( u ^{2} + 1 )} + \frac{1}{4 ( u + 1 )} - \frac{1}{4 ( u - 1 )} d u

Apply the Sum Rule:

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

= \frac{1}{2} (\int \frac{1}{2 ( u ^{2} + 1 )} d u + \int \frac{1}{4 ( u + 1 )} d u - \int \frac{1}{4 ( u - 1 )} d u)

\int \frac{1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} arctan (u)

\int \frac{1}{4 ( u + 1 )} d u = \frac{1}{4} ln ∣ u + 1 ∣

\int \frac{1}{4 ( u - 1 )} d u = \frac{1}{4} ln ∣ u - 1 ∣

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

Substitute back

u = t^{2}

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

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} arctan (t^{2}) + \frac{1}{4} ln t^{2} + 1 - \frac{1}{4} ln t^{2} - 1) + C

\frac{\partial}{\partial x} ((x - y)^{10})

x^{4}, x^{2}

f (x) = d e^{- c x}

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

f (t) = e^{3 t}

What is the integral of t/(1-t^8) ?
The integral of t/(1-t^8) is 1/2 (1/2 arctan(t^2)+1/4 ln|t^2+1|-1/4 ln|t^2-1|)+C