What is the integral of (9x+1)/(9x^2+2x-1) ?

The integral of (9x+1)/(9x^2+2x-1) is 1/2 ln|9x^2+2x-1|+C

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integral of (9x+1)/(9x^2+2x-1)

\int \frac{9 x + 1}{9 x ^{2} + 2 x - 1} d x

\frac{1}{2} ln 9 x^{2} + 2 x - 1 + C

Solution steps

\int \frac{9 x + 1}{9 x ^{2} + 2 x - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = 9 x^{2} + 2 x - 1

= \frac{1}{2} ln 9 x^{2} + 2 x - 1

Add a constant to the solution

= \frac{1}{2} ln 9 x^{2} + 2 x - 1 + C

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What is the integral of (9x+1)/(9x^2+2x-1) ?
The integral of (9x+1)/(9x^2+2x-1) is 1/2 ln|9x^2+2x-1|+C