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

The integral of x/(3-2x^2) is -1/4 ln|3-2x^2|+C

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integral of x/(3-2x^2)

\int \frac{x}{3 - 2 x ^{2}} d x

- \frac{1}{4} ln 3 - 2 x^{2} + C

Solution steps

\int \frac{x}{3 - 2 x ^{2}} d x

Apply u-substitution

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

= \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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= - \frac{1}{4} ln ∣ u ∣

Substitute back

u = 3 - 2 x^{2}

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

Add a constant to the solution

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

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

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\int \frac{1}{1 - \frac{x ^{2}}{36}} d x

What is the integral of x/(3-2x^2) ?
The integral of x/(3-2x^2) is -1/4 ln|3-2x^2|+C