What is the integral of (4x-13)/(2x^2+x-6) ?

The integral of (4x-13)/(2x^2+x-6) is ln|16x^2+8x-48|+2(ln| 4/7 x+8/7 |-ln| 4/7 x-6/7 |)+C

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integral of (4x-13)/(2x^2+x-6)

\int \frac{4 x - 13}{2 x ^{2} + x - 6} d x

ln 16 x^{2} + 8 x - 48 + 2 (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7}) + C

Solution steps

\int \frac{4 x - 13}{2 x ^{2} + x - 6} d x

Expand

\frac{4 x - 13}{2 x ^{2} + x - 6} : \frac{4 x}{2 x ^{2} + x - 6} - \frac{13}{2 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{4 x}{2 x ^{2} + x - 6} d x - \int \frac{13}{2 x ^{2} + x - 6} d x

\int \frac{4 x}{2 x ^{2} + x - 6} d x = ln 16 x^{2} + 8 x - 48 + \frac{1}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7})

\int \frac{13}{2 x ^{2} + x - 6} d x = - \frac{13}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7})

= ln 16 x^{2} + 8 x - 48 + \frac{1}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7}) - (- \frac{13}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7}))

Simplify

ln 16 x^{2} + 8 x - 48 + \frac{1}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7}) - (- \frac{13}{7} (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7})) : ln 16 x^{2} + 8 x - 48 + 2 (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7})

= ln 16 x^{2} + 8 x - 48 + 2 (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7})

Add a constant to the solution

= ln 16 x^{2} + 8 x - 48 + 2 (ln \frac{4}{7} x + \frac{8}{7} - ln \frac{4}{7} x - \frac{6}{7}) + C

(y - x 4) d x + (x + y 4) d y = 0

\frac{( 1 + y )}{( 1 + x )}

x^{2} - 1

\frac{d y}{d x} = 4 x e^{2 x}

\frac{\partial}{\partial x} (x + ln (5 x + y))

What is the integral of (4x-13)/(2x^2+x-6) ?
The integral of (4x-13)/(2x^2+x-6) is ln|16x^2+8x-48|+2(ln| 4/7 x+8/7 |-ln| 4/7 x-6/7 |)+C