What is the integral of x/((x^4+2)) ?

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

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integral of x/((x^4+2))

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

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

Solution steps

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

Apply u-substitution

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

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

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{2 ( 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}{2} \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}{2} arctan (v)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

t y^{^{'}} + 2 y = sin (t), y (\frac{π}{2}) = 7

x^{2} - 41, 7 - 2 x

\int sin (θ) sin (3 θ) sin (5 θ) d θ

x, at x = 100

2 t^{2}

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