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

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

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

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

6 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C

Solution steps

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

Expand

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

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

\int \frac{3 x}{x ^{2} + x - 6} d x = 6 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x))

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

= 6 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)

Add a constant to the solution

= 6 (\frac{1}{4} ln 4 x^{2} + 4 x - 24 + \frac{1}{20} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x)) - \frac{1}{5} (ln \frac{6}{5} + \frac{2}{5} x - ln - \frac{4}{5} + \frac{2}{5} x) + C

x = 7 y^{2}, x = 3 + 4 y^{2}

\int sin (mπ x) d x

f (x) = 2 x - \frac{e ^{x}}{2}

\int_{π}^{\infty} \frac{4}{x ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + y = (3 x + 4) e^{3 x}