What is the integral of (x+10)/(sqrt(2-x-3x^2)) ?

The integral of (x+10)/(sqrt(2-x-3x^2)) is 1/(6sqrt(3))(-sqrt(-36x^2-12x+24)+59arcsin((6x+1)/5))+C

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integral of (x+10)/(sqrt(2-x-3x^2))

\int \frac{x + 10}{2 - x - 3 x ^{2}} d x

\frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5})) + C

Solution steps

\int \frac{x + 10}{2 - x - 3 x ^{2}} d x

Apply u-substitution

: \int \frac{u}{- 3 u ^{2} + 59 u - 288} d u

= \int \frac{u}{- 3 u ^{2} + 59 u - 288} d u

Complete the square

- 3 u^{2} + 59 u - 288 : - 3 (u - \frac{59}{6})^{2} + \frac{25}{12}

= \int \frac{u}{- 3 ( u - \frac{59}{6} ) ^{2} + \frac{25}{12}} d u

Apply u-substitution

: \int \frac{6 v + 59}{3 - 36 v ^{2} + 25} d v

= \int \frac{6 v + 59}{3 - 36 v ^{2} + 25} d v

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{6 v + 59}{- 36 v ^{2} + 25} d v

Expand

\frac{6 v + 59}{- 36 v ^{2} + 25} : \frac{6 v}{- 36 v ^{2} + 25} + \frac{59}{- 36 v ^{2} + 25}

= \frac{1}{3} \cdot \int \frac{6 v}{- 36 v ^{2} + 25} + \frac{59}{- 36 v ^{2} + 25} d v

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}{3} (\int \frac{6 v}{- 36 v ^{2} + 25} d v + \int \frac{59}{- 36 v ^{2} + 25} d v)

\int \frac{6 v}{- 36 v ^{2} + 25} d v = - \frac{1}{6} - 36 v^{2} + 25

\int \frac{59}{- 36 v ^{2} + 25} d v = \frac{59}{6} arcsin (\frac{6}{5} v)

= \frac{1}{3} (- \frac{1}{6} - 36 v^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} v))

Substitute back

= \frac{1}{3} - \frac{1}{6} - 36 (x + 10 - \frac{59}{6})^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} (x + 10 - \frac{59}{6}))

Simplify

\frac{1}{3} - \frac{1}{6} - 36 (x + 10 - \frac{59}{6})^{2} + 25 + \frac{59}{6} arcsin (\frac{6}{5} (x + 10 - \frac{59}{6})) : \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5}))

= \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5}))

Add a constant to the solution

= \frac{1}{6 3} (- - 36 x^{2} - 12 x + 24 + 59 arcsin (\frac{6 x + 1}{5})) + C

\frac{d}{d x} (3 x^{2} - 9)

n = 0 \sum \infty \frac{4 ^{n}}{5 ^{n}}

f (x) = (3 x + 70)^{50}

6 x^{\frac{3}{2}}

y^{^{''}} + 6.2832 y = 0

What is the integral of (x+10)/(sqrt(2-x-3x^2)) ?
The integral of (x+10)/(sqrt(2-x-3x^2)) is 1/(6sqrt(3))(-sqrt(-36x^2-12x+24)+59arcsin((6x+1)/5))+C