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

f (x) = \frac{5}{x}, at x = 1

\int t e^{t} d t

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

y = \frac{1}{x ^{6}}, (1, 1)

\int \frac{4 x}{( 2 x ^{2} - 7 ) ^{10}} d x

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