What is the integral of 1/(sqrt(-x^2+8x-12)) ?

The integral of 1/(sqrt(-x^2+8x-12)) is arcsin(1/2 (x-4))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(sqrt(-x^2+8x-12))

\int \frac{1}{- x ^{2} + 8 x - 12} d x

arcsin (\frac{1}{2} (x - 4)) + C

Solution steps

\int \frac{1}{- x ^{2} + 8 x - 12} d x

Complete the square

- x^{2} + 8 x - 12 : - (x - 4)^{2} + 4

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

Apply u-substitution

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

Simplify

= arcsin (\frac{1}{2} (x - 4))

Add a constant to the solution

= arcsin (\frac{1}{2} (x - 4)) + C

What is the integral of 1/(sqrt(-x^2+8x-12)) ?
The integral of 1/(sqrt(-x^2+8x-12)) is arcsin(1/2 (x-4))+C