What is the integral of (ax)/(x^4+b^4) ?

The integral of (ax)/(x^4+b^4) is a 1/(2b^2)arctan((x^2)/(b^2))+C

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integral of (ax)/(x^4+b^4)

\int \frac{a x}{x ^{4} + b ^{4}} d x

a \frac{1}{2 b ^{2}} arctan (\frac{x ^{2}}{b ^{2}}) + C

Solution steps

\int \frac{a x}{x ^{4} + b ^{4}} d x

Take the constant out:

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

= a \cdot \int \frac{x}{x ^{4} + b ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( u ^{2} + b ^{4} )} d u

= a \cdot \int \frac{1}{2 ( u ^{2} + b ^{4} )} d u

Take the constant out:

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

= a \frac{1}{2} \cdot \int \frac{1}{u ^{2} + b ^{4}} d u

Apply Integral Substitution

= a \frac{1}{2} \cdot \int \frac{1}{b ^{2} ( v ^{2} + 1 )} d v

Take the constant out:

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

= a \frac{1}{2} \cdot \frac{1}{b ^{2}} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= a \frac{1}{2} \cdot \frac{1}{b ^{2}} arctan (v)

Substitute back

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

Simplify

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

= a \frac{1}{2 b ^{2}} arctan (\frac{x ^{2}}{b ^{2}})

Add a constant to the solution

= a \frac{1}{2 b ^{2}} arctan (\frac{x ^{2}}{b ^{2}}) + C

x \to 0 lim (\frac{- 1}{5 x})

f (x) = sec^{2} (2 x)

x \to - 2 lim (2 x + 2 + x)

\frac{\partial}{\partial x} (e^{x y z^{7}})

\frac{d}{d x} (x^{2} + sin (x))

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