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

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

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

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

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

Solution steps

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

Complete the square

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

\frac{\partial}{\partial x} (r + 2 r^{4} x + 4 x)

(4 + y^{2}) d x + (9 + x^{2}) d y = 0

\int \frac{x ^{4} + 2 x + 6}{x ^{3} + x ^{2} - 2 x} d x

\int \frac{11}{11 + e ^{x}} d x

\frac{d}{d t} (tan (e^{2 t}) + e^{t a n (2 t)})

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