What is the integral of t/(t^4+16) ?

The integral of t/(t^4+16) is 1/8 arctan((t^2)/4)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of t/(t^4+16)

\int \frac{t}{t ^{4} + 16} d t

\frac{1}{8} arctan (\frac{t ^{2}}{4}) + C

Solution steps

\int \frac{t}{t ^{4} + 16} d t

Apply u-substitution

= \int \frac{1}{2 ( u ^{2} + 16 )} 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} + 16} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{4 ( 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}{4} \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}{4} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{4} arctan (\frac{t ^{2}}{4})

Simplify

\frac{1}{2} \cdot \frac{1}{4} arctan (\frac{t ^{2}}{4}) : \frac{1}{8} arctan (\frac{t ^{2}}{4})

= \frac{1}{8} arctan (\frac{t ^{2}}{4})

Add a constant to the solution

= \frac{1}{8} arctan (\frac{t ^{2}}{4}) + C

What is the integral of t/(t^4+16) ?
The integral of t/(t^4+16) is 1/8 arctan((t^2)/4)+C