What is the integral of t/(t^4+9) ?

The integral of t/(t^4+9) is 1/6 arctan((t^2)/3)+C

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integral of t/(t^4+9)

\int \frac{t}{t ^{4} + 9} d t

\frac{1}{6} arctan (\frac{t ^{2}}{3}) + C

Solution steps

\int \frac{t}{t ^{4} + 9} d t

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + 9 )} d u

= \int \frac{1}{2 ( u ^{2} + 9 )} 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} + 9} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{3 ( 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}{3} \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}{3} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{3} arctan (\frac{t ^{2}}{3})

Simplify

\frac{1}{2} \cdot \frac{1}{3} arctan (\frac{t ^{2}}{3}) : \frac{1}{6} arctan (\frac{t ^{2}}{3})

= \frac{1}{6} arctan (\frac{t ^{2}}{3})

Add a constant to the solution

= \frac{1}{6} arctan (\frac{t ^{2}}{3}) + C

\frac{d}{d x} (4 x^{3} e^{x})

x \to - \infty lim (- 4 x)

\frac{\partial}{\partial x} (x^{2} y)

\frac{d}{d x} ((e^{2 x})^{s e c (x)})

f (t) = 2 sin (t)

What is the integral of t/(t^4+9) ?
The integral of t/(t^4+9) is 1/6 arctan((t^2)/3)+C