What is the integral of (8x)/(x^4+1) ?

The integral of (8x)/(x^4+1) is 4arctan(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 (8x)/(x^4+1)

\int \frac{8 x}{x ^{4} + 1} d x

4 arctan (x^{2}) + C

Solution steps

\int \frac{8 x}{x ^{4} + 1} d x

Take the constant out:

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

= 8 \cdot \int \frac{x}{x ^{4} + 1} d x

Apply u-substitution

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

= 8 \cdot \int \frac{1}{2 ( u ^{2} + 1 )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} arctan (u)

Substitute back

u = x^{2}

= 8 \cdot \frac{1}{2} arctan (x^{2})

Simplify

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

= 4 arctan (x^{2})

Add a constant to the solution

= 4 arctan (x^{2}) + C

\int \frac{x}{4 x ^{2} - 16} d x

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

\int \frac{1}{2 x ^{2} - 3 x + 9} d x

\int \frac{x}{1 - x ^{2}} arcsin (x) d x

\int 3 e^{- 3 t} d t

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