What is the integral of x/(x^4+6x^2+10) ?

The integral of x/(x^4+6x^2+10) is 1/2 arctan(x^2+3)+C

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integral of x/(x^4+6x^2+10)

\int \frac{x}{x ^{4} + 6 x ^{2} + 10} d x

\frac{1}{2} arctan (x^{2} + 3) + C

Solution steps

\int \frac{x}{x ^{4} + 6 x ^{2} + 10} d x

Apply u-substitution

: \int \frac{1}{4 u ( 6 u + u + 10 )} d u

= \int \frac{1}{4 u ( 6 u + u + 10 )} 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 ( 6 u + u + 10 )} d u

Apply u-substitution

: \int \frac{2}{6 v + v ^{2} + 10} d v

= \frac{1}{4} \cdot \int \frac{2}{6 v + v ^{2} + 10} 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}{6 v + v ^{2} + 10} d v

Complete the square

6 v + v^{2} + 10 : (v + 3)^{2} + 1

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{( v + 3 ) ^{2} + 1} d v

Apply u-substitution

: \int \frac{1}{w ^{2} + 1} d w

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

Substitute back

= \frac{1}{4} \cdot 2 arctan (x^{4} + 3)

Simplify

\frac{1}{4} \cdot 2 arctan (x^{4} + 3) : \frac{1}{2} arctan (x^{2} + 3)

= \frac{1}{2} arctan (x^{2} + 3)

Add a constant to the solution

= \frac{1}{2} arctan (x^{2} + 3) + C

\int \frac{1}{4 - ( u + 3 )} d u

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

f (x) = \frac{- 6 x}{x ^{2} + 1}, at x = 0

\int tan (θ) sec (θ) sec^{2} (θ) d θ

\frac{d}{d x} (ln (4 + ln (x)))

What is the integral of x/(x^4+6x^2+10) ?
The integral of x/(x^4+6x^2+10) is 1/2 arctan(x^2+3)+C