What is the integral of x/(16-6x-x^2) ?

The integral of x/(16-6x-x^2) is -1/2 ln|-x^2-6x+16|-3/10 (ln| 1/5 (x+3)+1|-ln| 1/5 (x+3)-1|)+C

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integral of x/(16-6x-x^2)

\int \frac{x}{16 - 6 x - x ^{2}} d x

- \frac{1}{2} ln - x^{2} - 6 x + 16 - \frac{3}{10} (ln \frac{1}{5} (x + 3) + 1 - ln \frac{1}{5} (x + 3) - 1) + C

Solution steps

\int \frac{x}{16 - 6 x - x ^{2}} d x

Complete the square

16 - 6 x - x^{2} : - (x + 3)^{2} + 25

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

Apply u-substitution

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

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

Expand

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

\int \frac{u}{- u ^{2} + 25} d u = - \frac{1}{2} ln - u^{2} + 25

\int \frac{3}{- u ^{2} + 25} d u = \frac{3}{10} (ln \frac{u}{5} + 1 - ln \frac{u}{5} - 1)

= - \frac{1}{2} ln - u^{2} + 25 - \frac{3}{10} (ln \frac{u}{5} + 1 - ln \frac{u}{5} - 1)

Substitute back

u = x + 3

= - \frac{1}{2} ln - (x + 3)^{2} + 25 - \frac{3}{10} (ln \frac{x + 3}{5} + 1 - ln \frac{x + 3}{5} - 1)

Simplify

- \frac{1}{2} ln - (x + 3)^{2} + 25 - \frac{3}{10} (ln \frac{x + 3}{5} + 1 - ln \frac{x + 3}{5} - 1) : - \frac{1}{2} ln - x^{2} - 6 x + 16 - \frac{3}{10} (ln \frac{1}{5} (x + 3) + 1 - ln \frac{1}{5} (x + 3) - 1)

= - \frac{1}{2} ln - x^{2} - 6 x + 16 - \frac{3}{10} (ln \frac{1}{5} (x + 3) + 1 - ln \frac{1}{5} (x + 3) - 1)

Add a constant to the solution

= - \frac{1}{2} ln - x^{2} - 6 x + 16 - \frac{3}{10} (ln \frac{1}{5} (x + 3) + 1 - ln \frac{1}{5} (x + 3) - 1) + C

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What is the integral of x/(16-6x-x^2) ?
The integral of x/(16-6x-x^2) is -1/2 ln|-x^2-6x+16|-3/10 (ln| 1/5 (x+3)+1|-ln| 1/5 (x+3)-1|)+C