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

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

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

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

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

Solution steps

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

= ln x^{2} + x - 3

Add a constant to the solution

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

\int \frac{ln ( x ^{5} )}{x} d x

\frac{2}{3} t^{2} + 5 e^{- 4 t}

x \to \infty lim (ln (3 x))

f (x) = \frac{16 x}{x ^{2} + 16}, at x = - 2

sin (x^{2})

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