What is the integral of x/(x^4+2x^2+5) ?

The integral of x/(x^4+2x^2+5) is 1/(2sqrt(7))arctan((x^2)/(sqrt(7)))+C

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integral of x/(x^4+2x^2+5)

\int \frac{x}{x ^{4} + 2 x ^{2} + 5} d x

\frac{1}{2 7} arctan (\frac{x ^{2}}{7}) + C

Solution steps

\int \frac{x}{x ^{4} + 2 x ^{2} + 5} d x

Apply u-substitution

: \int \frac{1}{4 u ( 2 u + u + 5 )} d u

= \int \frac{1}{4 u ( 2 u + u + 5 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ( 2 u + u + 5 )} d u

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{u ( 2 u + u + 5 )} = u^{- \frac{1}{2}} \frac{1}{( 2 + u + 5 )}

= \frac{1}{4} \cdot \int u^{- \frac{1}{2}} \frac{1}{2 + u + 5} d u

Simplify

u^{- \frac{1}{2}} \frac{1}{2 + u + 5} : \frac{1}{u ( u + 7 )}

= \frac{1}{4} \cdot \int \frac{1}{u ( u + 7 )} d u

Apply u-substitution

: \int \frac{2}{v ^{2} + 7} d v

= \frac{1}{4} \cdot \int \frac{2}{v ^{2} + 7} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{v ^{2} + 7} d v

Apply Integral Substitution

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{7 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{7} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

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

= \frac{1}{4} \cdot 2 \cdot \frac{1}{7} arctan (w)

Substitute back

= \frac{1}{4} \cdot 2 \cdot \frac{1}{7} arctan (\frac{x ^{4}}{7})

Simplify

\frac{1}{4} \cdot 2 \cdot \frac{1}{7} arctan (\frac{x ^{4}}{7}) : \frac{1}{2 7} arctan (\frac{x ^{2}}{7})

= \frac{1}{2 7} arctan (\frac{x ^{2}}{7})

Add a constant to the solution

= \frac{1}{2 7} arctan (\frac{x ^{2}}{7}) + C

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What is the integral of x/(x^4+2x^2+5) ?
The integral of x/(x^4+2x^2+5) is 1/(2sqrt(7))arctan((x^2)/(sqrt(7)))+C