What is the integral of (4x)/(2x^2+1) ?

The integral of (4x)/(2x^2+1) is ln|2x^2+1|+C

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integral of (4x)/(2x^2+1)

\int \frac{4 x}{2 x ^{2} + 1} d x

ln 2 x^{2} + 1 + C

Solution steps

\int \frac{4 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

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

Apply u-substitution

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

= 4 \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 4 \cdot \frac{1}{4} ln ∣ u ∣

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

= ln 2 x^{2} + 1

Add a constant to the solution

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

\frac{d y}{d x} + 6 y = e^{x}

\int x^{3} + x^{\frac{1}{3}} d x

\int 5 sin^{3} (2 x co) s^{3} 2 x d x

y^{^{''}} + 4 y = 4 sin (2 t)

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

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