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

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

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

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

ln x^{2} + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 1

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

Simplify

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

= ln x^{2} + 1

Add a constant to the solution

= ln x^{2} + 1 + C

\int sin (6)

\frac{\partial}{\partial u} (sin (u) cos (v))

\frac{d y}{d x} = e^{y} sin (x)

\frac{d}{d x} (\frac{x ^{2}}{6 + x})

f (x) = \frac{1}{x + 2}, at x = 3

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