What is the integral of x/(sqrt(1+6x-3x^2)) ?

The integral of x/(sqrt(1+6x-3x^2)) is -1/3 sqrt(-3x^2+6x+1)+1/(sqrt(3))arcsin((sqrt(3))/2 (x-1))+C

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integral of x/(sqrt(1+6x-3x^2))

\int \frac{x}{1 + 6 x - 3 x ^{2}} d x

- \frac{1}{3} - 3 x^{2} + 6 x + 1 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1)) + C

Solution steps

\int \frac{x}{1 + 6 x - 3 x ^{2}} d x

Complete the square

1 + 6 x - 3 x^{2} : - 3 (x - 1)^{2} + 4

= \int \frac{x}{- 3 ( x - 1 ) ^{2} + 4} d x

Apply u-substitution

= \int \frac{u + 1}{- 3 u ^{2} + 4} d u

Expand

\frac{u + 1}{- 3 u ^{2} + 4} : \frac{u}{- 3 u ^{2} + 4} + \frac{1}{- 3 u ^{2} + 4}

= \int \frac{u}{- 3 u ^{2} + 4} + \frac{1}{- 3 u ^{2} + 4} 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

= \int \frac{u}{- 3 u ^{2} + 4} d u + \int \frac{1}{- 3 u ^{2} + 4} d u

\int \frac{u}{- 3 u ^{2} + 4} d u = - \frac{1}{3} - 3 u^{2} + 4

\int \frac{1}{- 3 u ^{2} + 4} d u = \frac{1}{3} arcsin (\frac{3}{2} u)

= - \frac{1}{3} - 3 u^{2} + 4 + \frac{1}{3} arcsin (\frac{3}{2} u)

Substitute back

u = x - 1

= - \frac{1}{3} - 3 (x - 1)^{2} + 4 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1))

Simplify

- \frac{1}{3} - 3 (x - 1)^{2} + 4 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1)) : - \frac{1}{3} - 3 x^{2} + 6 x + 1 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1))

= - \frac{1}{3} - 3 x^{2} + 6 x + 1 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1))

Add a constant to the solution

= - \frac{1}{3} - 3 x^{2} + 6 x + 1 + \frac{1}{3} arcsin (\frac{3}{2} (x - 1)) + C

What is the integral of x/(sqrt(1+6x-3x^2)) ?
The integral of x/(sqrt(1+6x-3x^2)) is -1/3 sqrt(-3x^2+6x+1)+1/(sqrt(3))arcsin((sqrt(3))/2 (x-1))+C