What is the integral of t/(t^4+81) ?

The integral of t/(t^4+81) is 1/18 arctan((t^2)/9)+C

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integral of t/(t^4+81)

\int \frac{t}{t ^{4} + 81} d t

\frac{1}{18} arctan (\frac{t ^{2}}{9}) + C

Solution steps

\int \frac{t}{t ^{4} + 81} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 81 )} d u

= \int \frac{1}{2 ( u ^{2} + 81 )} 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} + 81} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{9 ( 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}{9} \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}{9} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{9} arctan (\frac{t ^{2}}{9})

Simplify

\frac{1}{2} \cdot \frac{1}{9} arctan (\frac{t ^{2}}{9}) : \frac{1}{18} arctan (\frac{t ^{2}}{9})

= \frac{1}{18} arctan (\frac{t ^{2}}{9})

Add a constant to the solution

= \frac{1}{18} arctan (\frac{t ^{2}}{9}) + C

f (x) = \frac{1}{1 + x}

4 y^{^{''}} - 5 y^{^{'}} = 0

\frac{1}{x x ^{2} - 1}, x = \frac{2}{2}, x = 2

(\frac{5}{x})^{^{'}}

\int sin (3 x) ln (cos (3 x)) d x

What is the integral of t/(t^4+81) ?
The integral of t/(t^4+81) is 1/18 arctan((t^2)/9)+C