What is the integral of 1/(sqrt(-x^2-6x+40)) ?

The integral of 1/(sqrt(-x^2-6x+40)) is arcsin(1/7 (x+3))+C

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integral of 1/(sqrt(-x^2-6x+40))

\int \frac{1}{- x ^{2} - 6 x + 40} d x

arcsin (\frac{1}{7} (x + 3)) + C

Solution steps

\int \frac{1}{- x ^{2} - 6 x + 40} d x

Complete the square

- x^{2} - 6 x + 40 : - (x + 3)^{2} + 49

= \int \frac{1}{- ( x + 3 ) ^{2} + 49} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 49} d u

= \int \frac{1}{- u ^{2} + 49} 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}{7} (x + 3))

Simplify

= arcsin (\frac{1}{7} (x + 3))

Add a constant to the solution

= arcsin (\frac{1}{7} (x + 3)) + C

x \to 0 lim (cot^{2} (x))

x \to 2 lim (\frac{1}{x - 2})

\int \frac{x ^{2} + 4 x + 1}{x ^{3} + x} d x

\frac{d}{d x} (\frac{( x ^{2} - 1 )}{x ^{2}})

\int - sin (x) cos^{2} (x) d x

What is the integral of 1/(sqrt(-x^2-6x+40)) ?
The integral of 1/(sqrt(-x^2-6x+40)) is arcsin(1/7 (x+3))+C