What is the integral of t/(t^4+49) ?

The integral of t/(t^4+49) is 1/14 arctan((t^2)/7)+C

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integral of t/(t^4+49)

\int \frac{t}{t ^{4} + 49} d t

\frac{1}{14} arctan (\frac{t ^{2}}{7}) + C

Solution steps

\int \frac{t}{t ^{4} + 49} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 49 )} d u

= \int \frac{1}{2 ( u ^{2} + 49 )} 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} + 49} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{7 ( 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}{7} \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}{7} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{7} arctan (\frac{t ^{2}}{7})

Simplify

\frac{1}{2} \cdot \frac{1}{7} arctan (\frac{t ^{2}}{7}) : \frac{1}{14} arctan (\frac{t ^{2}}{7})

= \frac{1}{14} arctan (\frac{t ^{2}}{7})

Add a constant to the solution

= \frac{1}{14} arctan (\frac{t ^{2}}{7}) + C

y^{^{''}} - 2 y^{^{'}} - 3 y = - 3 x e^{- x}

\int \frac{x}{x - 5} d x

x \to 0 lim (x^{3} cos (\frac{1}{x}))

f (x) = (\frac{x x - 1}{2})

m (r) = \frac{r ^{3} + 3 r + 2}{r ^{2} - 1}

What is the integral of t/(t^4+49) ?
The integral of t/(t^4+49) is 1/14 arctan((t^2)/7)+C