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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add similar elements:

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x e^{w})

x \to 2 - lim (\frac{3 x}{x - 2})

f (x) = (x^{2} + 3 x)^{2}

\int \frac{8}{x ^{2} x ^{2} + 16} d x

x \to \infty lim (81 x^{2} + x - 9 x)

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