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

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

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

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

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

Solution steps

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

= ln x^{2} + x - 2

Add a constant to the solution

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

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

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

x \to \infty lim (\frac{4 ^{x + 1} + 2 ^{x}}{4 ^{x}})

\frac{d}{d x} (- 7 ln (5 x^{2} - 11))

\int_{2}^{3} \frac{1}{- 7 + 16 x - 4 x ^{2}} d x

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