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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{3} d v

= \int \frac{1}{3} 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 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot 1 \cdot arcsin (\frac{3}{2} (x - 1))

Simplify

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

Add a constant to the solution

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

y^{^{'}} - 2 y = 3 e^{- 2 t}

\frac{d}{d x} (\frac{e ^{x^{2}}}{2})

(x + 1) \frac{d y}{d x} + (x + 2) y = 6 x e^{- x}

lo g_{10} (n)

\int_{0}^{\frac{1}{2}} arcsin (2 x) d x

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