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

Apply Trigonometric Substitution

= \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

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