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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of t/(9+t^4)

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{1}{2 ( 9 + u ^{2} )} 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}{9 + u ^{2}} 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 y}{d x} = (1 - x) y^{2}, y (0) = - \frac{1}{12}

\int cos^{3} (x s i) n^{3} x d x

\int 4 x (x^{2} + 27000)^{- \frac{2}{3}} d x

\frac{d y}{d x} - 2 e^{3 x} - 20 e^{2 x} = e^{3 x}

\int \frac{1}{x 1 + 16 x ^{2}} d x

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