What is the integral of t/(t^4+5) ?

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

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integral of t/(t^4+5)

\int \frac{t}{t ^{4} + 5} d t

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

Solution steps

\int \frac{t}{t ^{4} + 5} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 5 )} d u

= \int \frac{1}{2 ( u ^{2} + 5 )} 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} + 5} 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}{2 5} arctan (\frac{t ^{2}}{5})

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

Add a constant to the solution

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

8 sin (3 t) + 2 cos (2 t)

\int \frac{e ^{s i n (x)}}{sec ( x )} d x

f (x) = (3 x^{3} + 5) (x^{4} - 5 x)

\int \frac{x + 5}{x ^{3} + 2 x ^{2} - x - 2} d x

2 y^{^{'}} = y

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