What is the integral of (x+1)/(3x^2+6x-5) ?

The integral of (x+1)/(3x^2+6x-5) is 1/6 ln|3x^2+6x-5|+C

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integral of (x+1)/(3x^2+6x-5)

\int \frac{x + 1}{3 x ^{2} + 6 x - 5} d x

\frac{1}{6} ln 3 x^{2} + 6 x - 5 + C

Solution steps

\int \frac{x + 1}{3 x ^{2} + 6 x - 5} d x

Expand

\frac{x + 1}{3 x ^{2} + 6 x - 5} : \frac{x}{3 x ^{2} + 6 x - 5} + \frac{1}{3 x ^{2} + 6 x - 5}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{x}{3 x ^{2} + 6 x - 5} d x + \int \frac{1}{3 x ^{2} + 6 x - 5} d x

\int \frac{x}{3 x ^{2} + 6 x - 5} d x = \frac{1}{6} ln 3 x^{2} + 6 x - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

\int \frac{1}{3 x ^{2} + 6 x - 5} d x = - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

= \frac{1}{6} ln 3 x^{2} + 6 x - 5 + \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1)

Add similar elements:

\frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) - \frac{1}{4 6} (ln \frac{3}{2 2} (x + 1) + 1 - ln \frac{3}{2 2} (x + 1) - 1) = 0

= \frac{1}{6} ln 3 x^{2} + 6 x - 5

Add a constant to the solution

= \frac{1}{6} ln 3 x^{2} + 6 x - 5 + C

\int 2 x (x^{2} - 4)^{6} d x

\int \frac{x}{x ^{2} + 10 x + 50} d x

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

\int 4 x^{3} x^{2} + 1 d x

\int 12 + 6

What is the integral of (x+1)/(3x^2+6x-5) ?
The integral of (x+1)/(3x^2+6x-5) is 1/6 ln|3x^2+6x-5|+C