What is the integral of (t^2)/(1+t^4) ?

The integral of (t^2)/(1+t^4) is

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integral of (t^2)/(1+t^4)

\int \frac{t ^{2}}{1 + t ^{4}} d t

- \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)) + C

Solution steps

\int \frac{t ^{2}}{1 + t ^{4}} d t

Take the partial fraction of

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

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

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

\int \frac{t}{2 2 ( t ^{2} + 2 t + 1 )} d t = \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1))

\int \frac{t}{2 2 ( t ^{2} - 2 t + 1 )} d t = \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))

= - \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1))

Add a constant to the solution

= - \frac{1}{4 2} (ln 2 t^{2} + 22 t + 2 - 2 arctan (2 t + 1)) + \frac{1}{4 2} (ln 2 t^{2} - 22 t + 2 + 2 arctan (2 t - 1)) + C

\frac{\partial}{\partial x} (x^{2} - 3 y^{2})

\int_{0}^{1} \frac{54}{4 y - 1} d y

g (t) = arcsin (2 t)

\frac{\partial}{\partial x} (8 e^{x y + 7})

5 e^{5 e^{x} + x}