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

The integral of x/(2x^2+x-3) is 2(1/8 ln|16x^2+8x-24|+1/40 (ln| 6/5+4/5 x|-ln|-4/5+4/5 x|))+C

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

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

2 (\frac{1}{8} ln 16 x^{2} + 8 x - 24 + \frac{1}{40} (ln \frac{6}{5} + \frac{4}{5} x - ln - \frac{4}{5} + \frac{4}{5} x)) + C

Solution steps

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

Complete the square

2 x^{2} + x - 3 : 2 (x + \frac{1}{4})^{2} - \frac{25}{8}

= \int \frac{x}{2 ( x + \frac{1}{4} ) ^{2} - \frac{25}{8}} d x

Apply u-substitution

: \int \frac{2 ( 4 u - 1 )}{16 u ^{2} - 25} d u

= \int \frac{2 ( 4 u - 1 )}{16 u ^{2} - 25} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{4 u - 1}{16 u ^{2} - 25} d u

Expand

\frac{4 u - 1}{16 u ^{2} - 25} : \frac{4 u}{16 u ^{2} - 25} - \frac{1}{16 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

= 2 (\int \frac{4 u}{16 u ^{2} - 25} d u - \int \frac{1}{16 u ^{2} - 25} d u)

\int \frac{4 u}{16 u ^{2} - 25} d u = \frac{1}{8} ln 16 u^{2} - 25

\int \frac{1}{16 u ^{2} - 25} d u = - \frac{1}{40} (ln \frac{4}{5} u + 1 - ln \frac{4}{5} u - 1)

= 2 (\frac{1}{8} ln 16 u^{2} - 25 - (- \frac{1}{40} (ln \frac{4}{5} u + 1 - ln \frac{4}{5} u - 1)))

Substitute back

u = x + \frac{1}{4}

= 2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} - 25 - (- \frac{1}{40} (ln \frac{4}{5} (x + \frac{1}{4}) + 1 - ln \frac{4}{5} (x + \frac{1}{4}) - 1)))

Simplify

2 (\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} - 25 - (- \frac{1}{40} (ln \frac{4}{5} (x + \frac{1}{4}) + 1 - ln \frac{4}{5} (x + \frac{1}{4}) - 1))) : 2 (\frac{1}{8} ln 16 x^{2} + 8 x - 24 + \frac{1}{40} (ln \frac{6}{5} + \frac{4}{5} x - ln - \frac{4}{5} + \frac{4}{5} x))

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x - 24 + \frac{1}{40} (ln \frac{6}{5} + \frac{4}{5} x - ln - \frac{4}{5} + \frac{4}{5} x))

Add a constant to the solution

= 2 (\frac{1}{8} ln 16 x^{2} + 8 x - 24 + \frac{1}{40} (ln \frac{6}{5} + \frac{4}{5} x - ln - \frac{4}{5} + \frac{4}{5} x)) + C

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\int \frac{1}{6 x - 9 x ^{2}} d x

What is the integral of x/(2x^2+x-3) ?
The integral of x/(2x^2+x-3) is 2(1/8 ln|16x^2+8x-24|+1/40 (ln| 6/5+4/5 x|-ln|-4/5+4/5 x|))+C