What is the integral of x/(2x^2-5) ?

The integral of x/(2x^2-5) is 1/4 ln|2x^2-5|+C

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integral of x/(2x^2-5)

\int \frac{x}{2 x ^{2} - 5} d x

\frac{1}{4} ln 2 x^{2} - 5 + C

Solution steps

\int \frac{x}{2 x ^{2} - 5} 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 = 2 x^{2} - 5

= \frac{1}{4} ln 2 x^{2} - 5

Add a constant to the solution

= \frac{1}{4} ln 2 x^{2} - 5 + C

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

f (x) = cos (3 x) ln (x^{4})

\int \frac{3 tan ( θ ) - 4 cos ^{2} ( θ )}{cos ( θ )} d θ

x \to 0 + lim (\frac{ln ^{2} ( x )}{2})

\frac{\partial}{\partial y} (x^{2} + \frac{x}{y})

What is the integral of x/(2x^2-5) ?
The integral of x/(2x^2-5) is 1/4 ln|2x^2-5|+C