What is the integral of (x^2)/((15+6x-9x^2)) ?

The integral of (x^2)/((15+6x-9x^2)) is

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integral of (x^2)/((15+6x-9x^2))

\int \frac{x ^{2}}{( 15 + 6 x - 9 x ^{2} )} d x

\frac{1}{216} (- 24 (x - \frac{1}{3}) - 8 ln - 9 x^{2} + 6 x + 15 + 17 (ln \frac{3}{4} + \frac{3}{4} x - ln - \frac{5}{4} + \frac{3}{4} x)) + C

Solution steps

\int \frac{x ^{2}}{15 + 6 x - 9 x ^{2}} d x

Complete the square

15 + 6 x - 9 x^{2} : - 9 (x - \frac{1}{3})^{2} + 16

= \int \frac{x ^{2}}{- 9 ( x - \frac{1}{3} ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{( 3 u + 1 ) ^{2}}{9 ( - 9 u ^{2} + 16 )} d u

= \int \frac{( 3 u + 1 ) ^{2}}{9 ( - 9 u ^{2} + 16 )} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int \frac{( 3 u + 1 ) ^{2}}{- 9 u ^{2} + 16} d u

Long division

\frac{( 3 u + 1 ) ^{2}}{- 9 u ^{2} + 16} : - 1 + \frac{6 u + 17}{- 9 u ^{2} + 16}

= \frac{1}{9} \cdot \int - 1 + \frac{6 u + 17}{- 9 u ^{2} + 16} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{9} (- \int 1 d u + \int \frac{6 u + 17}{- 9 u ^{2} + 16} d u)

\int 1 d u = u

\int \frac{6 u + 17}{- 9 u ^{2} + 16} d u = - \frac{1}{3} ln - 9 u^{2} + 16 + \frac{17}{24} (ln \frac{3}{4} u + 1 - ln \frac{3}{4} u - 1)

= \frac{1}{9} (- u - \frac{1}{3} ln - 9 u^{2} + 16 + \frac{17}{24} (ln \frac{3}{4} u + 1 - ln \frac{3}{4} u - 1))

Substitute back

u = x - \frac{1}{3}

= \frac{1}{9} (- (x - \frac{1}{3}) - \frac{1}{3} ln - 9 (x - \frac{1}{3})^{2} + 16 + \frac{17}{24} (ln \frac{3}{4} (x - \frac{1}{3}) + 1 - ln \frac{3}{4} (x - \frac{1}{3}) - 1))

Simplify

\frac{1}{9} (- (x - \frac{1}{3}) - \frac{1}{3} ln - 9 (x - \frac{1}{3})^{2} + 16 + \frac{17}{24} (ln \frac{3}{4} (x - \frac{1}{3}) + 1 - ln \frac{3}{4} (x - \frac{1}{3}) - 1)) : \frac{1}{216} (- 24 (x - \frac{1}{3}) - 8 ln - 9 x^{2} + 6 x + 15 + 17 (ln \frac{3}{4} + \frac{3}{4} x - ln - \frac{5}{4} + \frac{3}{4} x))

= \frac{1}{216} (- 24 (x - \frac{1}{3}) - 8 ln - 9 x^{2} + 6 x + 15 + 17 (ln \frac{3}{4} + \frac{3}{4} x - ln - \frac{5}{4} + \frac{3}{4} x))

Add a constant to the solution

= \frac{1}{216} (- 24 (x - \frac{1}{3}) - 8 ln - 9 x^{2} + 6 x + 15 + 17 (ln \frac{3}{4} + \frac{3}{4} x - ln - \frac{5}{4} + \frac{3}{4} x)) + C

n = 1 \sum \infty \frac{1}{3 ^{\frac{1}{n}}}

x \to 3 lim (6 x - 7)

f (x) = x^{2} - 8, at x = 7

\frac{d}{d x} (8 x + 6)

f (x) = - 2 x + 5

What is the integral of (x^2)/((15+6x-9x^2)) ?
The integral of (x^2)/((15+6x-9x^2)) is