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

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

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integral of (8x)/((x^4+1))

integral

\frac{8 x}{( x ^{4} + 1 )}

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

= 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

in t e g r a l \frac{1}{( 1 + x ^{2} ) ^{3}}

in t e g r a l csc^{4} (x)

in t e g r a l \frac{x ^{3}}{( x ^{2} + 1 )}

in t e g r a l \frac{x ^{2}}{( 1 - x ^{2} ) ^{2}}

in t e g r a l e^{z} (z + 1)

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