What is the integral of 1/(sqrt(-x^2-4x+21)) ?

The integral of 1/(sqrt(-x^2-4x+21)) is arcsin(1/5 (x+2))+C

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integral of 1/(sqrt(-x^2-4x+21))

\int \frac{1}{- x ^{2} - 4 x + 21} d x

arcsin (\frac{1}{5} (x + 2)) + C

Solution steps

\int \frac{1}{- x ^{2} - 4 x + 21} d x

Complete the square

- x^{2} - 4 x + 21 : - (x + 2)^{2} + 25

= \int \frac{1}{- ( x + 2 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 25} d u

= \int \frac{1}{- u ^{2} + 25} 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}{5} (x + 2))

Simplify

= arcsin (\frac{1}{5} (x + 2))

Add a constant to the solution

= arcsin (\frac{1}{5} (x + 2)) + C

\int \frac{2 e ^{x}}{1 - e ^{2 x}} d x

9 x^{2} sin (x) tan (x)

\int a cos (t) d t

\frac{d y}{d x} = \frac{5 ( 3 - y ) ^{2}}{( 3 - x ) ^{2}}

\frac{d}{d x} (\frac{7}{10} x^{2} + 4^{3})

What is the integral of 1/(sqrt(-x^2-4x+21)) ?
The integral of 1/(sqrt(-x^2-4x+21)) is arcsin(1/5 (x+2))+C