What is the integral of 1/(sqrt(-x^2-10x)) ?

The integral of 1/(sqrt(-x^2-10x)) is arcsin(1/5 x+1)+C

integral of 1/(sqrt(-x^2-10x))

\int \frac{1}{- x ^{2} - 10 x} d x

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

Solution steps

\int \frac{1}{- x ^{2} - 10 x} d x

Complete the square

- x^{2} - 10 x : - (x + 5)^{2} + 25

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

Apply u-substitution

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

Simplify

1 \cdot arcsin (\frac{1}{5} (x + 5)) : arcsin (\frac{1}{5} x + 1)

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

Add a constant to the solution

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

\int ln (1 + 2 x) d x

\int (2 x y + x^{2} - 1) d y

ma c l a u r in 9 sin (4 x)

d er i v a t i v e w e^{w}

\frac{\partial}{\partial x} (\frac{x ^{\frac{5}{7}}}{e ^{x}})

What is the integral of 1/(sqrt(-x^2-10x)) ?
The integral of 1/(sqrt(-x^2-10x)) is arcsin(1/5 x+1)+C