What is the integral of (2x+1)/(x^2+x-6) ?

The integral of (2x+1)/(x^2+x-6) is ln|x^2+x-6|+C

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integral of (2x+1)/(x^2+x-6)

\int \frac{2 x + 1}{x ^{2} + x - 6} d x

ln x^{2} + x - 6 + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + x - 6} 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 - 6

= ln x^{2} + x - 6

Add a constant to the solution

= ln x^{2} + x - 6 + C

\frac{d}{d x} (ln (\frac{x + 1}{x - 1}))

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

\int_{0}^{\frac{1}{300}} 94 sin (225) d t

\int \frac{2}{t ^{3}} d t

What is the integral of (2x+1)/(x^2+x-6) ?
The integral of (2x+1)/(x^2+x-6) is ln|x^2+x-6|+C