What is the integral of (2x+6)/(x^2+6x-3) ?

The integral of (2x+6)/(x^2+6x-3) is ln|x^2+6x-3|+C

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integral of (2x+6)/(x^2+6x-3)

integral

\frac{2 x + 6}{x ^{2} + 6 x - 3}

ln x^{2} + 6 x - 3 + C

Solution steps

\int \frac{2 x + 6}{x ^{2} + 6 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} + 6 x - 3

= ln x^{2} + 6 x - 3

Add a constant to the solution

= ln x^{2} + 6 x - 3 + C

(e^{x})^{2} x^{2}

\frac{x ^{2}}{4 - x ^{2}}

\frac{1}{( 100 + x ^{2} )}

\frac{( r ^{2} )}{( 1 + r ^{2} ) d}

sin^{3} (2 x) cos (2 x)

What is the integral of (2x+6)/(x^2+6x-3) ?
The integral of (2x+6)/(x^2+6x-3) is ln|x^2+6x-3|+C