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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

Simplify

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

Add a constant to the solution

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

0.4241 (y - 25) + 70 x = - 10845.81 \cdot \frac{d y}{d x}

\int \frac{3 x ^{3} - 2 x ^{2} + x - 5}{x ^{2}} d x

t an g e n t f (x) = \frac{e ^{x}}{x ^{2}}, at x = 1

n = 1 \sum \infty \frac{3}{5 ^{n}}

\int \frac{3 x ^{5}}{1 - 12 x ^{6}} d x

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