What is the integral of 1/(sqrt(16+4x-2x^2)) ?

The integral of 1/(sqrt(16+4x-2x^2)) is 1/(sqrt(2))arcsin((x-1)/3)+C

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integral of 1/(sqrt(16+4x-2x^2))

\int \frac{1}{16 + 4 x - 2 x ^{2}} d x

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

Solution steps

\int \frac{1}{16 + 4 x - 2 x ^{2}} d x

Complete the square

16 + 4 x - 2 x^{2} : - 2 (x - 1)^{2} + 18

= \int \frac{1}{- 2 ( x - 1 ) ^{2} + 18} d x

Apply u-substitution

: \int \frac{1}{- 2 u ^{2} + 18} d u

= \int \frac{1}{- 2 u ^{2} + 18} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= \int \frac{1}{2} d v

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int_{1}^{6} 2 π x (- \frac{3}{2} x + 9) d x

\frac{x + 1}{x + 2}

\frac{d}{d x} (y - x^{2})

\int \frac{48 x}{4 x ^{2} + e} d x

\int_{- 1}^{1} \frac{x}{x ^{2} + 1} d x

What is the integral of 1/(sqrt(16+4x-2x^2)) ?
The integral of 1/(sqrt(16+4x-2x^2)) is 1/(sqrt(2))arcsin((x-1)/3)+C