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

The integral of 1/(sqrt(6+x-x^2)) is arcsin(2/5 x-1/5)+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(6+x-x^2))

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

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

Solution steps

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

Complete the square

6 + x - x^{2} : - (x - \frac{1}{2})^{2} + \frac{25}{4}

= \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{25}{4}} d x

Apply u-substitution

: \int \frac{2}{- 4 u ^{2} + 25} d u

= \int \frac{2}{- 4 u ^{2} + 25} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{1}{- 4 u ^{2} + 25} d u

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{2} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{5} (x - \frac{1}{2})) : arcsin (\frac{2}{5} x - \frac{1}{5})

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

Add a constant to the solution

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

\int \frac{1}{2 + 9 x ^{2}} d x

\int x cos (6 π x^{2}) d x

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

\int x x + 15 d x

\int \frac{5 x ln ( x )}{x ^{2} - 4} d x

What is the integral of 1/(sqrt(6+x-x^2)) ?
The integral of 1/(sqrt(6+x-x^2)) is arcsin(2/5 x-1/5)+C