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

The integral of x/(sqrt(5+4x-x^2)) is -sqrt(-x^2+4x+5)+2arcsin(1/3 (x-2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(sqrt(5+4x-x^2))

\int \frac{x}{5 + 4 x - x ^{2}} d x

- - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)) + C

Solution steps

\int \frac{x}{5 + 4 x - x ^{2}} d x

Complete the square

5 + 4 x - x^{2} : - (x - 2)^{2} + 9

= \int \frac{x}{- ( x - 2 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{u + 2}{- u ^{2} + 9} d u

= \int \frac{u + 2}{- u ^{2} + 9} d u

Apply Trigonometric Substitution

: \int 3 sin (v) + 2 d v

= \int 3 sin (v) + 2 d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 3 sin (v) d v + \int 2 d v

\int 3 sin (v) d v = - 3 cos (v)

\int 2 d v = 2 v

= - 3 cos (v) + 2 v

Substitute back

= - 3 cos (arcsin (\frac{1}{3} (x - 2))) + 2 arcsin (\frac{1}{3} (x - 2))

Simplify

- 3 cos (arcsin (\frac{1}{3} (x - 2))) + 2 arcsin (\frac{1}{3} (x - 2)) : - - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2))

= - - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2))

Add a constant to the solution

= - - x^{2} + 4 x + 5 + 2 arcsin (\frac{1}{3} (x - 2)) + C

\int_{1}^{\infty} x^{2} d x

f (x) = \frac{2 x ^{2} + 6 x + 6}{x + 1}

\frac{d y}{d x} = e^{4 y} sin (5 x)

\frac{\partial}{\partial x} (6 x + 4 y)

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

What is the integral of x/(sqrt(5+4x-x^2)) ?
The integral of x/(sqrt(5+4x-x^2)) is -sqrt(-x^2+4x+5)+2arcsin(1/3 (x-2))+C