What is the integral of t/(t^4+25) ?

The integral of t/(t^4+25) is 1/10 arctan((t^2)/5)+C

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integral of t/(t^4+25)

\int \frac{t}{t ^{4} + 25} d t

\frac{1}{10} arctan (\frac{t ^{2}}{5}) + C

Solution steps

\int \frac{t}{t ^{4} + 25} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 25 )} d u

= \int \frac{1}{2 ( u ^{2} + 25 )} 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} + 25} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{5 ( 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}{5} \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}{5} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{5} arctan (\frac{t ^{2}}{5})

Simplify

\frac{1}{2} \cdot \frac{1}{5} arctan (\frac{t ^{2}}{5}) : \frac{1}{10} arctan (\frac{t ^{2}}{5})

= \frac{1}{10} arctan (\frac{t ^{2}}{5})

Add a constant to the solution

= \frac{1}{10} arctan (\frac{t ^{2}}{5}) + C

\int \frac{2 x + 1}{x ^{2} + x - 3} d x

\int \frac{ln ( x ^{5} )}{x} d x

\frac{2}{3} t^{2} + 5 e^{- 4 t}

x \to \infty lim (ln (3 x))

f (x) = \frac{16 x}{x ^{2} + 16}, at x = - 2

What is the integral of t/(t^4+25) ?
The integral of t/(t^4+25) is 1/10 arctan((t^2)/5)+C