What is the integral of (2x)/(2x^2+1) ?

The integral of (2x)/(2x^2+1) is 1/2 ln|2x^2+1|+C

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integral of (2x)/(2x^2+1)

\int \frac{2 x}{2 x ^{2} + 1} d x

\frac{1}{2} ln 2 x^{2} + 1 + C

Solution steps

\int \frac{2 x}{2 x ^{2} + 1} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{2 x ^{2} + 1} d x

Apply u-substitution

: \int \frac{1}{4 u} d u

= 2 \cdot \int \frac{1}{4 u} d u

Take the constant out:

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

= 2 \cdot \frac{1}{4} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} + 1

= 2 \cdot \frac{1}{4} ln 2 x^{2} + 1

Simplify

2 \cdot \frac{1}{4} ln 2 x^{2} + 1 : \frac{1}{2} ln 2 x^{2} + 1

= \frac{1}{2} ln 2 x^{2} + 1

Add a constant to the solution

= \frac{1}{2} ln 2 x^{2} + 1 + C

\frac{d y}{d x} = \frac{3 x ^{2}}{e ^{2 y}}, y (0) = \frac{1}{2}

2 x (x^{2} - 4 x + 5)^{8}, at x = 2

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 t^{3}

(\frac{2}{x ^{3}})^{^{'}}

y = \frac{1}{3} x^{3}

What is the integral of (2x)/(2x^2+1) ?
The integral of (2x)/(2x^2+1) is 1/2 ln|2x^2+1|+C