What is the integral of 1/(sqrt(-x^2-8x)) ?

The integral of 1/(sqrt(-x^2-8x)) is arcsin(1/4 x+1)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(-x^2-8x))

\int \frac{1}{- x ^{2} - 8 x} d x

arcsin (\frac{1}{4} x + 1) + C

Solution steps

\int \frac{1}{- x ^{2} - 8 x} d x

Complete the square

- x^{2} - 8 x : - (x + 4)^{2} + 16

= \int \frac{1}{- ( x + 4 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{1}{- u ^{2} + 16} d u

= \int \frac{1}{- u ^{2} + 16} 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}{4} (x + 4))

Simplify

1 \cdot arcsin (\frac{1}{4} (x + 4)) : arcsin (\frac{1}{4} x + 1)

= arcsin (\frac{1}{4} x + 1)

Add a constant to the solution

= arcsin (\frac{1}{4} x + 1) + C

\int \frac{d}{1} (x sin (x)) d x

\int \frac{x + 3}{x} d x

\int - x^{2} + 2 x + 8 d x

\int (20 x^{4} + 8) (2 x^{5} + 4 x)^{15} d x

\int 80 x^{\frac{1}{2}} d x

What is the integral of 1/(sqrt(-x^2-8x)) ?
The integral of 1/(sqrt(-x^2-8x)) is arcsin(1/4 x+1)+C