What is the integral of (9x)/(1+x^4) ?

The integral of (9x)/(1+x^4) is 9/2 arctan(x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (9x)/(1+x^4)

\int \frac{9 x}{1 + x ^{4}} d x

\frac{9}{2} arctan (x^{2}) + C

Solution steps

\int \frac{9 x}{1 + x ^{4}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{x}{1 + x ^{4}} d x

Apply u-substitution

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

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

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

= 9 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x^{2}

= 9 \cdot \frac{1}{2} arctan (x^{2})

Simplify

9 \cdot \frac{1}{2} arctan (x^{2}) : \frac{9}{2} arctan (x^{2})

= \frac{9}{2} arctan (x^{2})

Add a constant to the solution

= \frac{9}{2} arctan (x^{2}) + C

x \to 2 - lim (x + 5)

\frac{5 x}{x + 3}

\int 2 + x d x

\frac{d}{d x} (\frac{25 - x ^{2}}{16 x})

\frac{\partial ^{2}}{\partial x \partial y} (1 + x y^{2})

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