What is the integral of (4x)/(2x^2+3) ?

The integral of (4x)/(2x^2+3) is ln|2x^2+3|+C

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integral of (4x)/(2x^2+3)

\int \frac{4 x}{2 x ^{2} + 3} d x

ln 2 x^{2} + 3 + C

Solution steps

\int \frac{4 x}{2 x ^{2} + 3} 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} + 3} 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} + 3

= 4 \cdot \frac{1}{4} ln 2 x^{2} + 3

Simplify

4 \cdot \frac{1}{4} ln 2 x^{2} + 3 : ln 2 x^{2} + 3

= ln 2 x^{2} + 3

Add a constant to the solution

= ln 2 x^{2} + 3 + C

\int \frac{7}{x ^{2} + 4} d x

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

\frac{d}{d x} (- arcsin (\frac{a ^{2} - x ^{2}}{a}))

y = x^{2}, at x = - \frac{3}{4}

\frac{d}{d x} (ln ∣ csc (x) - cot (x) ∣)

What is the integral of (4x)/(2x^2+3) ?
The integral of (4x)/(2x^2+3) is ln|2x^2+3|+C