What is the integral of (2x+3)/(x^2+3x-5) ?

The integral of (2x+3)/(x^2+3x-5) is ln|x^2+3x-5|+C

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integral of (2x+3)/(x^2+3x-5)

\int \frac{2 x + 3}{x ^{2} + 3 x - 5} d x

ln x^{2} + 3 x - 5 + C

Solution steps

\int \frac{2 x + 3}{x ^{2} + 3 x - 5} 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} + 3 x - 5

= ln x^{2} + 3 x - 5

Add a constant to the solution

= ln x^{2} + 3 x - 5 + C

\int sin (x) (e^{c o s (x)}) d x

\int \frac{21 t - 12}{t} d t

\int \frac{1}{3 t} d t

\int x (x + 3)^{5} d x

\int x 49 + x^{2} d x

What is the integral of (2x+3)/(x^2+3x-5) ?
The integral of (2x+3)/(x^2+3x-5) is ln|x^2+3x-5|+C