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

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

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integral of t/(t^4+2)

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{2} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} arctan (\frac{t ^{2}}{2})

Simplify

\frac{1}{2} \cdot \frac{1}{2} arctan (\frac{t ^{2}}{2}) : \frac{1}{2 2} arctan (\frac{t ^{2}}{2})

= \frac{1}{2 2} arctan (\frac{t ^{2}}{2})

Add a constant to the solution

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

\frac{- 4 s + 8}{s ^{2} + 6 s + 8}

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What is the integral of t/(t^4+2) ?
The integral of t/(t^4+2) is 1/(2sqrt(2))arctan((t^2)/(sqrt(2)))+C