What is the integral of 1/(sqrt(5+4x-x^2)) ?

The integral of 1/(sqrt(5+4x-x^2)) is arcsin(1/3 (x-2))+C

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integral of 1/(sqrt(5+4x-x^2))

\int \frac{1}{5 + 4 x - x ^{2}} d x

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

Solution steps

\int \frac{1}{5 + 4 x - x ^{2}} d x

Complete the square

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

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

Apply u-substitution

: \int \frac{1}{- u ^{2} + 9} d u

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

Simplify

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

Add a constant to the solution

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

x \to - \infty lim (\frac{( 4 x ^{2} + 1 )}{( 2 - 7 x )})

x y^{^{'}} + 4 y = - 25 x sin (x^{5})

\frac{d}{d x} (\frac{3}{7 x ^{4} - 5 x ^{2} + 1})

\frac{d y}{d x} = tan (4 x)

\int 16 - t^{2} d t

What is the integral of 1/(sqrt(5+4x-x^2)) ?
The integral of 1/(sqrt(5+4x-x^2)) is arcsin(1/3 (x-2))+C