What is the integral of x/(3x^2+8x-3) ?

The integral of x/(3x^2+8x-3) is 1/6 ln|9x^2+24x-9|+2/15 (ln| 9/5+3/5 x|-ln|-1/5+3/5 x|)+C

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integral of x/(3x^2+8x-3)

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

\frac{1}{6} ln 9 x^{2} + 24 x - 9 + \frac{2}{15} (ln \frac{9}{5} + \frac{3}{5} x - ln - \frac{1}{5} + \frac{3}{5} x) + C

Solution steps

\int \frac{x}{3 x ^{2} + 8 x - 3} d x

Complete the square

3 x^{2} + 8 x - 3 : 3 (x + \frac{4}{3})^{2} - \frac{25}{3}

= \int \frac{x}{3 ( x + \frac{4}{3} ) ^{2} - \frac{25}{3}} d x

Apply u-substitution

: \int \frac{3 u - 4}{9 u ^{2} - 25} d u

= \int \frac{3 u - 4}{9 u ^{2} - 25} d u

Expand

\frac{3 u - 4}{9 u ^{2} - 25} : \frac{3 u}{9 u ^{2} - 25} - \frac{4}{9 u ^{2} - 25}

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 u}{9 u ^{2} - 25} d u - \int \frac{4}{9 u ^{2} - 25} d u

\int \frac{3 u}{9 u ^{2} - 25} d u = \frac{1}{6} ln 9 u^{2} - 25

\int \frac{4}{9 u ^{2} - 25} d u = - \frac{2}{15} (ln \frac{3}{5} u + 1 - ln \frac{3}{5} u - 1)

= \frac{1}{6} ln 9 u^{2} - 25 - (- \frac{2}{15} (ln \frac{3}{5} u + 1 - ln \frac{3}{5} u - 1))

Substitute back

u = x + \frac{4}{3}

= \frac{1}{6} ln 9 (x + \frac{4}{3})^{2} - 25 - (- \frac{2}{15} (ln \frac{3}{5} (x + \frac{4}{3}) + 1 - ln \frac{3}{5} (x + \frac{4}{3}) - 1))

Simplify

\frac{1}{6} ln 9 (x + \frac{4}{3})^{2} - 25 - (- \frac{2}{15} (ln \frac{3}{5} (x + \frac{4}{3}) + 1 - ln \frac{3}{5} (x + \frac{4}{3}) - 1)) : \frac{1}{6} ln 9 x^{2} + 24 x - 9 + \frac{2}{15} (ln \frac{9}{5} + \frac{3}{5} x - ln - \frac{1}{5} + \frac{3}{5} x)

= \frac{1}{6} ln 9 x^{2} + 24 x - 9 + \frac{2}{15} (ln \frac{9}{5} + \frac{3}{5} x - ln - \frac{1}{5} + \frac{3}{5} x)

Add a constant to the solution

= \frac{1}{6} ln 9 x^{2} + 24 x - 9 + \frac{2}{15} (ln \frac{9}{5} + \frac{3}{5} x - ln - \frac{1}{5} + \frac{3}{5} x) + C

x^{^{''}} + 3 x^{^{'}} + 2 x = 0

x + 4

\frac{d y}{d x} + \frac{6 y}{x} = x

x \to 4 lim (arctan (\frac{x ^{2} - 16}{7 x ^{2} - 28 x}))

\int \frac{1}{( 2 x + 1 )} d x

What is the integral of x/(3x^2+8x-3) ?
The integral of x/(3x^2+8x-3) is 1/6 ln|9x^2+24x-9|+2/15 (ln| 9/5+3/5 x|-ln|-1/5+3/5 x|)+C