What is the integral of 1/(sqrt(10+4x-x^2)) ?

The integral of 1/(sqrt(10+4x-x^2)) is arcsin(1/(sqrt(14))(x-2))+C

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integral of 1/(sqrt(10+4x-x^2))

\int \frac{1}{10 + 4 x - x ^{2}} d x

arcsin (\frac{1}{14} (x - 2)) + C

Solution steps

\int \frac{1}{10 + 4 x - x ^{2}} d x

Complete the square

10 + 4 x - x^{2} : - (x - 2)^{2} + 14

= \int \frac{1}{- ( x - 2 ) ^{2} + 14} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 14} d u

= \int \frac{1}{- u ^{2} + 14} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

= 1 \cdot arcsin (\frac{1}{14} (x - 2))

Simplify

= arcsin (\frac{1}{14} (x - 2))

Add a constant to the solution

= arcsin (\frac{1}{14} (x - 2)) + C

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

x y^{^{'}} + 3 y = \frac{( 2 e ^{- 2 x} )}{x ^{2}}

x \to \infty lim (7 + \frac{1}{x})

x^{3} + y^{3} = 9 x y, (4, 2)

What is the integral of 1/(sqrt(10+4x-x^2)) ?
The integral of 1/(sqrt(10+4x-x^2)) is arcsin(1/(sqrt(14))(x-2))+C