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

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

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integral of (4x+1)/(2x^2+x-3)

\int \frac{4 x + 1}{2 x ^{2} + x - 3} d x

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

Solution steps

\int \frac{4 x + 1}{2 x ^{2} + x - 3} d x

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = 2 x^{2} + x - 3

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

Add a constant to the solution

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

\frac{4}{s ( s ^{2} + 4 )}

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

\frac{\partial}{\partial x} (\frac{x ^{2} + t x}{sin ( t x )})

f (x) = \frac{x}{x ^{2} + 9}, at x = 0

y = 3 x^{2} - 4

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