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

The integral of 1/(sqrt(3x-2x^2)) is 1/(sqrt(2))arcsin(4/3 x-1)+C

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

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

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

Solution steps

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

Complete the square

3 x - 2 x^{2} : - 2 (x - \frac{3}{4})^{2} + \frac{9}{8}

= \int \frac{1}{- 2 ( x - \frac{3}{4} ) ^{2} + \frac{9}{8}} d x

Apply u-substitution

: \int \frac{2 2}{- 16 u ^{2} + 9} d u

= \int \frac{2 2}{- 16 u ^{2} + 9} d u

Take the constant out:

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

= 22 \cdot \int \frac{1}{- 16 u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int \frac{1}{4} d v

= 22 \cdot \int \frac{1}{4} d v

Take the constant out:

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

= 22 \frac{1}{4} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 22 \frac{1}{4} \cdot 1 \cdot v

Substitute back

= 22 \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (x - \frac{3}{4}))

Simplify

22 \frac{1}{4} \cdot 1 \cdot arcsin (\frac{4}{3} (x - \frac{3}{4})) : \frac{1}{2} arcsin (\frac{4}{3} x - 1)

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

Add a constant to the solution

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

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What is the integral of 1/(sqrt(3x-2x^2)) ?
The integral of 1/(sqrt(3x-2x^2)) is 1/(sqrt(2))arcsin(4/3 x-1)+C