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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int \frac{2}{4 v + v ^{2} + 5} d v

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

Complete the square

4 v + v^{2} + 5 : (v + 2)^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

x \to 2 + lim (\frac{2 - x}{∣ x - 2 ∣})

t \to 0 lim (\frac{e ^{\frac{1}{t}}}{t})

\frac{d y}{d x}, 2 x^{3} - 5 x y - y^{2} = 3

x \to - 2 lim ((6 x + 1) (2 x - 3))

\frac{d}{d x} (5^{l n (c o s (3 x))})

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